Learning Objectives

  • Recognize when to apply L’Hôpital’s rule.
  • Identify indeterminate forms produced by quotients, products, differences, and powers, and apply L’Hôpital’s rule in each case.
  • Describe the relative growth rates of functions.

In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule, uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value.

Applying L’Hôpital’s Rule

L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider \ds{\lim_{x\to a}}\,\frac{f(x)}{g(x)}.

If \underset{x\to a}{\text{lim}}f(x)={L}_{1}\text{ and }\underset{x\to a}{\text{lim}}g(x)={L}_{2}\ne 0, then \ds \underset{x\to a}{\text{lim}}\,\frac{f(x)}{g(x)}=\frac{{L}_{1}}{{L}_{2}}.

However, what happens if \underset{x\to a}{\text{lim}}f(x)=0 and \underset{x\to a}{\text{lim}}g(x)=0? In this case, we have to deal with what’s called the indeterminate form of type \ds\frac{0}{0}. This is considered an indeterminate form because we cannot determine the exact behavior of the quotient \ds\frac{f(x)}{g(x)} as x approaches a without further analysis. For example, consider the limits \ds\underset{x\to 2}{\text{lim}}\,\frac{{x}^{2}-4}{x-2}\text{ and }\underset{x\to 0}{\text{lim}}\,\frac{ \sin (x)}{x}.

The first limit can be evaluated by factoring the numerator:

\ds\underset{x\to 2}{\text{lim}}\,\frac{{x}^{2}-4}{x-2}=\underset{x\to 2}{\text{lim}}\,\frac{(x+2)(x-2)}{x-2}=\underset{x\to 2}{\text{lim}}(x+2)=2+2=4.

As for \ds\underset{x\to 0}{\text{lim}}\,\frac{ \sin (x)}{x}, one can use a geometric argument to show that \ds\underset{x\to 0}{\text{lim}}\,\frac{ \sin (x)}{x}=1.
Note that although both limits we just considered exist, in fact, anything is possible for the indeterminate form: the limit might exist and be equal to any real number L, the limit might not exist, but there might be a trend of \infty or -\infty, or there might be no limit and no trend. For example, knowing that \ds\underset{x\to 0}{\text{lim}}\,\frac{ \sin (x)}{x}=1, it is easy to show that \ds\underset{x\to 0}{\text{lim}}\,\frac{ \sin (x)}{x^3}=\infty.

We are going to develop a universal technique for evaluating limits such as the ones above. Not only does it allow for an alternative and sometimes easier way to evaluate these limits, but also, and more importantly, it works for evaluating many other limits that we could not calculate before.

The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions \ds f and \ds g such that \ds\underset{x\to a}{\text{lim}}f(x)=0=\underset{x\to a}{\text{lim}}g(x) and \ds{g}^{\prime }(a)\ne 0. For \ds x that is very close to \ds a, we can write \ds f(x)\approx f(a)+{f}^{\prime }(a)(x-a) and \ds g(x)\approx g(a)+{g}^{\prime }(a)(x-a). (This follows either from the geometric understanding that the tangent line is very close to the graph around the point where the tangent “touches” the curve or, algebraically, from the formal limit definition of the derivative.)

Therefore, \ds\,\frac{f(x)}{g(x)}\approx \,\frac{f(a)+{f}^{\prime }(a)(x-a)}{g(a)+{g}^{\prime }(a)(x-a)}.

Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f’(a)(x – a) and y = g(a) + g’(a)(x – a) are also drawn.
Figure 1. If \ds\underset{x\to a}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}g(x), then the ratio \ds f(x)\text{/}g(x) is approximately equal to the ratio of their linear approximations near \ds a.

Since \ds f is differentiable at \ds a, it is continuous at \ds a, and therefore \ds f(a)=\underset{x\to a}{\text{lim}}f(x)=0. Similarly, \ds g(a)=\underset{x\to a}{\text{lim}}g(x)=0. If we also assume that \ds{f}^{\prime } and \ds{g}^{\prime } are continuous at \ds x=a, then \ds{f}^{\prime }(a)=\underset{x\to a}{\text{lim}}{f}^{\prime }(x) and \ds{g}^{\prime }(a)=\underset{x\to a}{\text{lim}}{g}^{\prime }(x). Using these arguments, we conclude that

\ds\underset{x\to a}{\text{lim}}\,\frac{f(x)}{g(x)}=\underset{x\to a}{\text{lim}}\,\frac{{f}^{\prime }(x)(x-a)}{{g}^{\prime }(x)(x-a)}=\underset{x\to a}{\text{lim}}\,\frac{{f}^{\prime }(x)}{{g}^{\prime }(x)}.

Note that the assumptions that \ds{f}^{\prime } and \ds{g}^{\prime } are continuous at \ds a and \ds{g}^{\prime }(a)\ne 0 can be loosened. We state L’Hôpital’s rule formally for the indeterminate form \ds\,\frac{0}{0}. Also note that writing \ds\,\frac{0}{0} does not mean we are actually dividing zero by zero. It is just the notation we use when dealing with a limit of a quotient with both numerator and denominator approaching zero.

L’Hôpital’s Rule ( \ds \frac00 Case)

Suppose \ds\,f and \ds\,g are differentiable functions over an open interval containing \ds\,a, except possibly at \ds\,a. If \ds\,\underset{x\to a}{\text{lim}}f(x)=0 and \ds\,\underset{x\to a}{\text{lim}}g(x)=0, then

\ds\,\underset{x\to a}{\text{lim}}\,\frac{f(x)}{g(x)}=\underset{x\to a}{\text{lim}}\,\frac{{f}^{\prime }(x)}{{g}^{\prime }(x)}, provided that the limit on the right exists or has a trend of \ds\,\infty or \ds\,-\infty . This result also holds for one-sided limits, or if a finite number a is replaced with \ds\,\infty \text{ or }-\infty .

Proof

We provide a proof of this theorem in the special case when f,g,{f}^{\prime }, and {g}^{\prime } are all continuous over an open interval containing a. In this case, since \underset{x\to a}{\text{lim}}f(x)=0=\underset{x\to a}{\text{lim}}g(x) and f and g are continuous at a, we have that f(a)=0=g(a). Therefore,

\begin{array}{ccc}\hfill \underset{x\to a}{\text{lim}}\ds\,\frac{f(x)}{g(x)}& =\underset{x\to a}{\text{lim}}\ds\,\frac{f(x)-f(a)}{g(x)-g(a)}\hfill & \quad\text{since}\ f(a)=0=g(a)\hfill \\[5mm] & =\underset{x\to a}{\text{lim}}\ds\,\frac{\ds\,\frac{f(x)-f(a)}{x-a}}{\ds\,\frac{g(x)-g(a)}{x-a}}\hfill & \quad \text{algebra}\hfill\\[5mm] & =\displaystyle \frac{\underset{x\to a}{\text{lim}}\ds\,\frac{f(x)-f(a)}{x-a}}{\underset{x\to a}{\text{lim}}\ds\,\frac{g(x)-g(a)}{x-a}}\hfill & \quad \text{limit of a quotient}\hfill\\[5mm] & \ds=\,\frac{{f}^{\prime }(a)}{{g}^{\prime }(a)}\hfill & \quad \text{definition of the derivative}\hfill \\[5mm] & =\ds\,\frac{\underset{x\to a}{\text{lim}}{f}^{\prime }(x)}{\underset{x\to a}{\text{lim}}{g}^{\prime }(x)}\hfill & \quad \text{continuity of}\  {f}^{\prime }\text{ and }{g}^{\prime }\hfill \\[5mm] & =\underset{x\to a}{\text{lim}}\ds\,\frac{{f}^{\prime }(x)}{{g}^{\prime }(x)}.\hfill & \quad \text{limit of a quotient}\hfill \end{array}

Note that L’Hôpital’s rule allows to calculate the limit of a quotient \ds\,\frac{f}{g} by considering the limit of the quotient of the derivatives \ds\,\frac{{f}^{\prime }}{{g}^{\prime }}. It is important to realize that we are not calculating the derivative of the quotient \ds\,\frac{f}{g}.

Applying L’Hôpital’s Rule ( \ds\frac00 Case)

Evaluate each of the following limits using L’Hôpital’s rule.

  1. \underset{x\to 0}{\text{lim}}\ds\,\frac{1- \cos (x)}{x}

  2. \underset{x\to 1}{\text{lim}}\ds\,\frac{ \sin (\pi x)}{\text{ln}(x)}

  3. \underset{x\to \infty }{\text{lim}}\ds\,\frac{{e}^{1\text{/}x}-1}{1\text{/}x}

  4. \underset{x\to 0}{\text{lim}}\ds\,\frac{ \sin (x)-x}{{x}^{2}}

Solution

  1. Since the numerator \big(1- \cos (x)\big) approaches 1-\cos(0)=0 and the denominator x approaches 0 when x\to 0, we can apply L’Hôpital’s rule to evaluate this limit. We have

    \begin{array}{cc}\hfill \underset{x\to 0}{\text{lim}}\ds\,\frac{1- \cos (x)}{x}& =\underset{x\to 0}{\text{lim}}\ds\,\frac{\ds\,\big(1- \cos (x)\big)'}{\ds\,(x)'}=\underset{x\to 0}{\text{lim}}\ds\,\frac{ \sin (x)}{1}\hfill \\[5mm] \hfill &=\ds\,\frac{\underset{x\to 0}{\text{lim}}\,\sin (x)}{\underset{x\to 0}{\text{lim}}1}=\,\frac{\sin(0)}1=\ds\,\frac{0}{1}=0.\hfill \end{array}
  2. As x\to 1,  \sin (\pi x)\to \sin(\pi)=0 and \text{ln}(x)\to \ln(1)=0. Therefore, we can apply L’Hôpital’s rule. We obtain

    \begin{array}{cc}\hfill \underset{x\to 1}{\text{lim}}\ds\,\frac{ \sin (\pi x)}{\text{ln}(x)}&=\underset{x\to 1}{\text{lim}}\ds\,\frac{ \big(\sin (\pi x)\big)'}{\big(\text{ln}(x)\big)'} =\underset{x\to 1}{\text{lim}}\ds\,\frac{\cos (\pi x)\cdot \pi}{1\text{/}x}\hfill \\[5mm] & =\ds\frac\,{\cos(\pi)\cdot\pi}{1/1}=(-1)\pi=-\pi .\hfill \end{array}

  3. When x\to \infty , we have that \dfrac1x\to0, that is, the denominator approaches 0, and the numerator ({e}^{1\text{/}x}-1}) approaches e^0-1=0. Therefore, we can apply L’Hôpital’s rule. We obtain

    \underset{x\to \infty }{\text{lim}}\ds\,\frac{{e}^{1\text{/}x}-1}{1\text{/}x}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{{e}^{1\text{/}x}\left(\ds\,\frac{-1}{{x}^{2}}\right)}{\ds\,\frac{-1}{{x}^{2}}}=\underset{x\to \infty }{\text{lim}}{e}^{1\text{/}x}={e}^{0}=1.

  4. As x\to 0, both numerator and denominator approach zero: \big(\sin(x)-x\big)\to\sin(0)-0=0 and x^2\to 0^2=0. Therefore, we can apply L’Hôpital’s rule. We obtain

    \underset{x\to 0}{\text{lim}}\ds\,\frac{ \sin (x)-x}{{x}^{2}}=\underset{x\to 0}{\text{lim}}\ds\,\frac{ \cos (x)-1}{2x}.

    Since numerator and denominator of this new quotient both approach 0 as x\to 0 \Big(\big(\cos(x)-1\big)\to \cos(0)-1=0 and 2x\to2\cdot0=0\Big), we apply L’Hôpital’s rule again. In doing so, we see that

    \underset{x\to 0}{\text{lim}}\ds\,\frac{ \cos (x)-1}{2x}=\underset{x\to 0}{\text{lim}}\ds\,\frac{- \sin (x)}{2}=\dfrac{-\sin(0)}2=0.

    Therefore, we conclude that

    \underset{x\to 0}{\text{lim}}\ds\,\frac{ \sin (x)-x}{{x}^{2}}=0.

Evaluate \underset{x\to 0}{\text{lim}}\ds\,\frac{x^3-x}{ \tan (x)}.

Answer

-1

We can also use L’Hôpital’s rule to evaluate limits of quotients \ds\,\frac{f(x)}{g(x)} in which f(x)\to \pm \infty and g(x)\to \pm \infty . Limits of this form are classified as indeterminate forms of type \ds \frac{\infty}{\infty} . Again, note that we are not actually dividing \infty by \infty . Since \infty is not a real number, that is impossible; rather, \ds \frac{\infty}{\infty} is the notation we use when dealing with a limit of a quotient with both numerator and denominator having an infinite trend of \infty or -\infty.

L’Hôpital’s Rule ( \ds\frac{\infty}{\infty} Case)

Suppose that f and g are differentiable functions over an open interval containing a, except possibly at a. Further suppose that \big(\underset{x\to a}{\text{lim}}f(x)=\infty\  \text{or}\  \underset{x\to a}{\text{lim}}f(x)=-\infty\big) and \big(\underset{x\to a}{\text{lim}}g(x)=\infty\  \text{or}\  \underset{x\to a}{\text{lim}}g(x)=-\infty\big). Then,

\underset{x\to a}{\text{lim}}\ds\,\frac{f(x)}{g(x)}=\underset{x\to a}{\text{lim}}\ds\,\frac{{f}^{\prime }(x)}{{g}^{\prime }(x)}, provided that the limit on the right exists or has a trend of \infty or -\infty . This result also holds for one-sided limits or if the finite number a is replaced with \infty or -\infty.

Applying L’Hôpital’s Rule ( \ds\frac{\infty}{\infty} Case)

Evaluate each of the following limits using L’Hôpital’s rule.

  1. \underset{x\to \infty }{\text{lim}}\ds\,\frac{3x+5}{2x+1}

  2. \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{\text{ln}(x)}{ \cot (x)}

Solution

  1. Since 3x+5 and 2x+1 are first-degree polynomials with positive leading coefficients, \underset{x\to \infty }{\text{lim}}(3x+5)=\infty and \underset{x\to \infty }{\text{lim}}(2x+1)=\infty . Therefore, we apply L’Hôpital’s rule and obtain

    \underset{x\to \infty }{\text{lim}}\ds\,\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{3}{2}=\ds\,\frac{3}{2}.

    Note that this limit can also be calculated without invoking L’Hôpital’s rule, by dividing the numerator and denominator by the highest power of x in the denominator:

    \underset{x\to \infty }{\text{lim}}\ds\,\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{3+5\text{/}x}{2+1\text{/}x}=\frac{3+0}{2+0}=\ds\,\frac{3}{2}.

    L’Hôpital’s rule provides us with an alternative means of evaluating limits of this type.

  2. Here, \underset{x\to {0}^{+}}{\text{lim}}\text{ln}(x)=-\infty and \underset{x\to {0}^{+}}{\text{lim}} \cot (x)=\infty . Therefore, we can apply L’Hôpital’s rule and obtain

    \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{\text{ln}(x)}{ \cot (x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{1\text{/}x}{-{ \csc }^{2}(x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{1}{-x\cdot{ \csc }^{2}(x)}.

    Now, as x\to {0}^{+}, { \csc }^{2}(x)\to \infty . Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In this case, anything can happen to the product, and we cannot make any conclusion yet. (Actually, this is an indeterminate form 0\cdot\infty that will be discussed a bit later.) To evaluate the limit, we use the relation of  \csc (x)=\dfrac1{\sin(x)} to rewrite the quotient under the limit:

    \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{1}{-x\cdot{ \csc }^{2}(x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{1}{\frac{-x}{\sin^2(x)}}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{{ \sin }^{2}(x)}{-x}.

    Now, \underset{x\to {0}^{+}}{\text{lim}}{ \sin }^{2}(x)=\sin^2(0)=0 and \underset{x\to {0}^{+}}{\text{lim}}(-x)=0, so we can apply L’Hôpital’s rule again:

    \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{{ \sin }^{2}(x)}{-x}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{2 \sin (x) \cos (x)}{-1}=\frac{2\sin(0)\cos(0)}{-1}=\ds\,\frac{0}{-1}=0.

    We conclude that \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{\text{ln}(x)}{ \cot (x)}=0.

Evaluate \underset{x\to \infty }{\text{lim}}\ds\,\frac{\text{ln}(x)}{5x}.

Answer

0

As was already mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L’Hôpital’s rule to a quotient \ds\,\frac{f(x)}{g(x)}, it is essential that the limit of \ds\,\frac{f(x)}{g(x)} be of the form \ds\,\frac{0}{0} or \ds\frac{\infty}{\infty} . This will be illustrated in the example below.

When L’Hôpital’s Rule Does Not Apply

Consider \ds\underset{x\to 1}{\text{lim}}\ds\,\frac{{x}^{2}+5}{3x+4}. Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

Solution

Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we tried to do so, we would have erroneously concluded that

\underset{x\to 1}{\text{lim}}\ds\,\frac{{x}^{2}+5}{3x+4}=\underset{x\to 1}{\text{lim}}\ds\,\frac{\big(x^2+5\big)'}{\big(3x+4\big)'}=\underset{x\to 1}{\text{lim}}\ds\,\frac{2x}{3}=\ds\,\frac{2}{3}.

However, since \underset{x\to 1}{\text{lim}}({x}^{2}+5)=6 and \underset{x\to 1}{\text{lim}}(3x+4)=7\ne0, using the arithmetic properties of limits, we actually have that \underset{x\to 1}{\text{lim}}\ds\,\frac{{x}^{2}+5}{3x+4}=\ds\,\frac{6}{7}, and so that

\underset{x\to 1}{\text{lim}}\ds\,\frac{{x}^{2}+5}{3x+4}\ne \underset{x\to 1}{\text{lim}}\ds\,\frac{\ds\,\big({x}^{2}+5\big)'}{\ds\,\big(3x+4\big)'}.

Another, trickier, case when L’Hôpital’s rule cannot be applied is when the limit of the quotient obtained after differentiation does not exist. Indeed, if we carefully read the statements of the L’Hôpital’s rule theorems, we can notice that they both say “provided that the limit on the right \left[\text{of}\ \dfrac{f'(x)}{g'(x)}\right] exists”. We explore the situation when this condition fails in the next example.

When L’Hôpital’s Rule Does Not Apply

Consider \underset{x\to \infty}{\text{lim}}\ds\,\frac{x+\sin(x)}{x}. Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

Solution

Both numerator and denominator approach infinity. Thus it appears that we can apply L’Hôpital’s rule. If we tried to do so, we would have erroneously concluded that

\underset{x\to \infty}{\text{lim}}\ds\,\frac{x+\sin(x)}{x}=\underset{x\to \infty}{\text{lim}}\ds\,\frac{\big(x+\sin(x)\big)'}{(x)'}=\underset{x\to \infty}{\text{lim}}\big((1+\cos(x)\big).

However, since \underset{x\to \infty}{\text{lim}}(1+\cos(x)) neither exists, nor is infinite, L’Hôpital’s rule does not actually apply.

Note that this does not mean that the original limit fails to exist. Because -1\le \sin(x)\le 1, when x>0, we have that

\ds\,\frac{x-1}{x}\le\frac{x+\sin(x)}{x}\le \frac{x+1}{x}. Now since

\ds \lim\limits_{x\to\infty}\,\frac{x-1}{x}=\lim\limits_{x\to\infty}\,\frac{x+1}{x}=1 (that can be proved using L’Hôpital’s rule), we can use the squeeze theorem to conclude that \underset{x\to \infty}{\text{lim}}\ds\,\frac{x+\sin(x)}{x} = 1.

Explain why we cannot apply L’Hôpital’s rule to evaluate \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{ \cos (x)}{x}. Evaluate \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{ \cos (x)}{x} by other means.

Answer

\infty.

Hint

Determine the limits of the numerator and denominator and analyze the behavior of the quotient.

Other Indeterminate Forms

We have seen that L’Hôpital’s rule is very useful for dealing with the indeterminate forms \dfrac{0}{0} and \ds\dfrac{\infty}{\infty}. It can also help to evaluate limits involving other indeterminate forms such as 0\cdot\infty, \infty -\infty , {1}^{\infty }, {\infty }^{0}, and {0}^{0}. As before, these expressions should be treated not as algebraic operations but as the notation reflecting the behavior of the function under the limit. We show why they are indeterminate forms and how to use L’Hôpital’s rule to evaluate the corresponding limits. The key idea is to rewrite the expression as a quotient of the form \dfrac{0}{0} or \ds\frac{\infty}{\infty}.

Indeterminate Form of Type 0\cdot\infty

Suppose we want to evaluate \underset{x\to a}{\text{lim}}(f(x)\cdot g(x)), where f(x)\to 0 and g(x)\to \pm\infty as x\to a. Since one term in the product is approaching zero but the other term is becoming arbitrarily large in magnitude, anything can happen to the product. We use the notation 0\cdot\infty to denote the form that arises in this situation. The expression 0\cdot\infty is considered indeterminate because, without further analysis, we cannot determine the exact behavior of the product f(x)g(x) as x\to a.

For example, let n be a positive integer and consider \ds f(x)=\frac{1}{{x}^{n}+1}\text{ and }g(x)=3{x}^{2}. As x\to \infty, f(x)\to 0 and g(x)\to \infty. However, the limit of \ds f(x)g(x)=\frac{3{x}^{2}}{{x}^{n}+1} as x\to \infty varies, depending on n. Indeed,

if n=1, then \underset{x\to \infty }{\text{lim}}f(x)g(x)=\lim\limits_{x\to\infty}\,\dfrac{3x^2}{x+1}=\infty,

if n=2, then \underset{x\to \infty }{\text{lim}}f(x)g(x)=\lim\limits_{x\to\infty}\,\dfrac{3x^2}{x^2+1}=3,

and if n=3, then \underset{x\to \infty }{\text{lim}}f(x)g(x)=\lim\limits_{x\to\infty}\,\dfrac{3x^2}{x^3+1}=0. (All these limits can be evaluated by either dividing both numerator and denominator by the highest power of x in the denominator or by using L’Hôpital’s rule, multiple times, when needed).

We now consider another limit involving the indeterminate form 0\cdot\infty and show how to rewrite the function as a quotient to use L’Hôpital’s rule.

Indeterminate Form of Type 0\cdot\infty

Evaluate \underset{x\to {0}^{+}}{\text{lim}}x\,\text{ln}(x).

Solution

To be able to use L’Hôpital’s rule, we rewrite the function x\,\text{ln}(x) as a quotient in the following way: x\,\text{ln}(x)=\ds\,\frac{\text{ln}(x)}{1\text{/}x}.  Since \text{ln}(x)\to -\infty as x\to {0}^{+} and \ds\,\frac{1}{x}\to \infty as x\to {0}^{+}, we can apply L’Hôpital’s rule to obtain

\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{\text{ln}(x)}{1\text{/}x}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{\ds\,\big(\text{ln}(x)\big)'}{\ds\,(1\text{/}x)'}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{1\text{/}{x}}{-1\text{/}x^2}=\underset{x\to {0}^{+}}{\text{lim}}(-x)=0.

We conclude that \underset{x\to {0}^{+}}{\text{lim}}x\,\text{ln}(x)=0.

In general, one of the ways to transform a product of functions f(x)g(x) into a quotient is to use double reciprocals, that is, rewrite \ds f(x)g(x) as \dfrac{f(x)}{1\text{/}{g(x)}} or \ds \frac{g(x)}{1\text{/}{f(x)}}.

Evaluate \underset{x\to 0}{\text{lim}}\,x \cot (x).

Answer

1

Indeterminate Form of Type \infty -\infty

Suppose that \ds\lim_{x\to a} f(x)=\lim_{x\to a} g(x) =\infty (or \lim_{x\to a} f(x)=\lim_{x\to a} g(x) =-\infty). Then the limit \underset{x\to a }{\text{lim}}(f(x)-g(x)) represents an indeterminate form \infty -\infty.

To show that this is, indeed, an indeterminate form, and the answer could be anything, consider the following example. Let n be a positive integer, f(x)=3{x}^{n} and g(x)=3{x}^{2}+5. As x\to \infty, f(x)\to \infty and g(x)\to \infty, but \ds\lim_{x\to\infty} (f(x)-g(x)) depends on the value of the exponent n. To see this, we recall that, when x\to\pm\infty, the behavior of a polynomial P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0 (a_n\ne0) is determined by the leading term a_nx^n since P(x)=a_nx^n\left(1+\dfrac{a_{n-1}}{a_n}\cdot\dfrac1x+\ldots+\dfrac{a_0}{a_n}\cdot\dfrac1{x^n}\right) and \lim\limits_{x\to\pm\infty}\,\dfrac1{x^k}=0 for any k>0.

Therefore, if n=3, then \underset{x\to \infty }{\text{lim}}(f(x)-g(x))=\underset{x\to \infty }{\text{lim}}(3{x}^{3}-3{x}^{2}-5)=\infty.

On the other hand, if n=2, then \underset{x\to \infty }{\text{lim}}(f(x)-g(x))=\underset{x\to \infty }{\text{lim}}(3{x}^{2}-3{x}^{2}-5)=-5.

Finally, if n=1, then \underset{x\to \infty }{\text{lim}}(f(x)-g(x))=\underset{x\to \infty }{\text{lim}}(3x-3{x}^{2}-5)=\underset{x\to \infty }{\text{lim}}(-3{x}^{2}+3x-5)=-\infty.

In the next example, we show how to rewrite an expression involving the indeterminate form \infty -\infty as a fraction in order to apply L’Hôpital’s rule.

Indeterminate Form of Type \infty -\infty

Evaluate \underset{x\to {0}^{+}}{\text{lim}}\left(\ds\,\frac{1}{{x}^{2}}-\ds\,\frac{1}{ \tan (x)}\right).

Solution

Bringing the expression to a common denominator, we obtain

\ds\,\frac{1}{{x}^{2}}-\ds\,\frac{1}{ \tan (x)}=\ds\,\frac{\tan (x)-{x}^{2}}{{x}^{2} \tan (x)}.

Since  \lim\limits_{x\to0^+}\big(\tan (x)-{x}^{2}\big)=\tan(0)-0^2=0 and \lim\limits_{x\to0^+} {x}^{2} \tan (x)= 0^2\cdot\tan(0)=0, we can apply L’Hôpital’s rule.

\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{ \tan (x)-{x}^{2}}{{x}^{2} \tan (x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{ \big(\tan (x)-{x}^{2}\big)'}{\big({x}^{2} \tan (x)\big)'}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{{ \sec }^{2}(x)-2x}{2x \tan (x)+{x}^{2}{ \sec }^{2}(x)}.

We have that \lim\limits_{x\to0^+}\big({ \sec }^{2}(x)-2x\big)=\sec^2(0)-2\cdot0=1 and \lim\limits_{x\to0^+}\big(2x \tan(x)+{x}^{2}{ \sec }^{2}(x)\big)=2\cdot0\cdot\tan(0)+0^2\cdot\sec^(0)=0. Because the denominator is positive as x approaches zero from the right, we conclude that \underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{{ \sec }^{2}(x)-2x}{{x}^{2}{ \sec }^{2}(x)+2x \tan (x)}=\infty .

Therefore, \underset{x\to {0}^{+}}{\text{lim}}\left(\ds\,\frac{1}{{x}^{2}}-\ds\,\frac{1}{ \tan (x)}\right)=\infty .

In general, bringing the difference of fractions to a common denominator often works to reduce an \infty-\infty indeterminate form to \dfrac00 or \dfrac{\infty}{\infty} form, after which L’Hôpital’s rule can be applied.

Evaluate \underset{x\to {0}^{+}}{\text{lim}}\left(\ds\,\frac{1}{x}-\ds\,\frac{1}{ \sin (x)}\right).

Answer

0

Hint

Rewrite the difference of fractions as a single fraction.

Another types of indeterminate forms that arise when evaluating limits involve exponents. The expressions {0}^{0}, {\infty }^{0}, and {1}^{\infty } are all indeterminate forms. Again, these expressions are not meant to be evaluated using algebraic operations, rather they provide a notation used to describe the behavior of a function under the limit. We now demonstrate how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate forms.

Suppose we want to evaluate \underset{x\to a}{\text{lim}}f{(x)}^{g(x)} and we arrive at one of the indeterminate forms listed above. We proceed as follows. Let y=f{(x)}^{g(x)}. Then, \text{ln}(y)=\text{ln}\big(f{(x)}^{g(x)}\big)=g(x)\text{ln}(f(x)), and we first evaluate \ds\lim_{x\to a} \ln(y)=\lim_{x\to a}g(x)\ln(f(x)). This allows to reduce an indeterminate power to an indeterminate product. Indeed,  {0}^{0} corresponds to 0\cdot(-\infty), \infty^0 corresponds  to 0\cdot\infty, and 1^{\infty} corresponds to \infty\cdot0. We then use the techniques discussed earlier to rewrite the expression g(x)\text{ln}(f(x)) as a quotient so that we can apply L’Hôpital’s rule. Suppose that \underset{x\to a}{\text{lim}}g(x)\text{ln}(f(x))=L, where L may be \infty or -\infty . So we have that \underset{x\to a}{\text{lim}}\text{ln}(y)=L, and since the natural logarithm function is continuous, we conclude that \text{ln}(\underset{x\to a}{\text{lim}}y)=L. It follows that \underset{x\to a}{\text{lim}}f{(x)}^{g(x)}=\underset{x\to a}{\text{lim}}y=e^{\text{ln}\left(\underset{x\to a}{\text{lim}}y\right)}={e}^{L}.

Indeterminate Form of Type {\infty }^{0}

Evaluate \underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}.

Solution

Let y={x}^{1\text{/}x}. Then \ds\ln(y)=\text{ln}({x}^{1\text{/}x})=\ds\,\frac{1}{x}\text{ln}(x)=\ds\,\frac{\text{ln}(x)}{x}.

We first need to evaluate \ds\lim_{x\to\infty} \ln(y)=\underset{x\to \infty }{\text{lim}}\ds\,\frac{\text{ln}(x)}{x}. Since both numerator and denominator approach infinity as x\to\infty, we can apply L’Hôpital’s rule to obtain \underset{x\to \infty }{\text{lim}}\ds\,\frac{\text{ln}(x)}{x}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{1\text{/}x}{1}=0.

So we have that \underset{x\to \infty }{\text{lim}}\left[\text{ln}(y)\right]=0 and since the natural logarithm function is continuous, we conclude that \text{ln}\big(\underset{x\to \infty }{\text{lim}}y\big)=0.

This leads to \underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}=\underset{x\to \infty }{\text{lim}}y=e^{\text{ln}\big(\underset{x\to \infty }{\text{lim}}y\big)}={e}^{0}=1.

Evaluate \underset{x\to \infty }{\text{lim}}{x}^{1\text{/}\text{ln}(x)}.

Answer

e

Hint

Take y={x}^{1\text{/}\text{ln}(x)} and consider \ds\lim_{x\to\infty} \ln(y).

Indeterminate Form of Type {0}^{0}

Evaluate \underset{x\to {0}^{+}}{\text{lim}}{x}^{ \sin (x)}.

Solution

Let y={x}^{ \sin (x)}. Then \text{ln}(y)=\text{ln}({x}^{ \sin (x)})= \sin (x)\text{ln}(x).

We now evaluate \underset{x\to {0}^{+}}{\text{lim}} \sin (x)\text{ln}(x). Since \underset{x\to {0}^{+}}{\text{lim}} \sin (x)=0 and \underset{x\to {0}^{+}}{\text{lim}}\text{ln}(x)=-\infty , we have the indeterminate form 0\cdot\infty . To apply L’Hôpital’s rule, we need to rewrite  \sin (x)\text{ln}(x) as a fraction. We could write

 \sin (x)\text{ln}(x)=\ds\,\frac{ \sin (x)}{1\text{/}\text{ln}(x)} or  \sin (x)\text{ln}(x)=\ds\,\frac{\text{ln}(x)}{1\text{/} \sin (x)}=\ds\,\frac{\text{ln}(x)}{ \csc (x)}.

Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain

\begin{array}{ll}\ds\underset{x\to {0}^{+}}{\text{lim}} \sin (x)\text{ln}(x)&=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{ \sin (x)}{1\text{/}\text{ln}(x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{ \cos (x)}{\big(-1\text{/}{(\text{ln}(x))}^{2}\big)\cdot1\text{/}x}\\[5mm]&=\underset{x\to {0}^{+}}{\text{lim}}(-x{(\text{ln}(x))}^{2} \cos (x)).\end{array}

Unfortunately, we not only have another expression involving the indeterminate form 0\cdot\infty , but the new limit is even more complicated to evaluate than the one we started with. Instead, we try the second option. By writing  \sin (x)\text{ln}(x)=\ds\,\frac{\text{ln}(x)}{1\text{/} \sin (x)}=\ds\,\frac{\text{ln}(x)}{ \csc (x)}, and applying L’Hôpital’s rule, we obtain

\underset{x\to {0}^{+}}{\text{lim}} \sin (x)\text{ln}(x)=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{\text{ln}(x)}{ \csc (x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{1\text{/}x}{- \csc (x) \cot (x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{-1}{x \csc (x) \cot (x)}.

Using the fact that  \csc (x)=\ds\,\frac{1}{ \sin (x)} and  \cot (x)=\ds\,\frac{1}{ \tan (x)}, we can rewrite the last limit as

\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{-1}{x\cdot 1/\sin(x)\cdot 1/\tan(x)}=\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{-{ \sin }(x)\tan(x)}{x}.

Because \underset{x\to {0}^{+}}{\text{lim}}\ds\, -\sin(x)\tan(x)=-\sin(0)\tan(0)=0=\underset{x\to {0}^{+}}{\text{lim}}\ds\,x, we can apply L’Hôpital’s rule again to get

\begin{array}{ll}\underset{x\to {0}^{+}}{\text{lim}}\ds\,\frac{-{ \sin }(x)\tan(x)}{x}&=\lim\limits_{x\to0^+}\,\dfrac{-\cos(x)\tan(x)-\sin(x)\sec^2(x)}{1}\\[5mm]&=\dfrac{-1\cdot0-0\cdot1}{1}=0.\end{array}

We conclude that \underset{x\to {0}^{+}}{\text{lim}}\text{ln}(y)=0. Using the continuity of logarithmic function, we obtain that \text{ln}(\underset{x\to {0}^{+}}{\text{lim}}y)=0, and hence

\underset{x\to {0}^{+}}{\text{lim}}{x}^{ \sin (x)}=\underset{x\to {0}^{+}}{\text{lim}}y=e^{\text{ln}(\underset{x\to {0}^{+}}{\text{lim}}y)}={e}^{0}=1.

Evaluate \underset{x\to {0}^{+}}{\text{lim}}{x}^{x}.

Answer

1

Hint

Take y={x}^{1\text{/}\text{ln}(x)} and consider \ds\lim_{x\to0^+} \left[\ln(y)\right].

Growth Rates of Functions

Suppose the functions f and g both approach infinity as x\to \infty . Although the values of both functions become arbitrarily large as the values of x become sufficiently large, sometimes one function is growing more quickly than the other. For example, f(x)={x}^{2} and g(x)={x}^{3} both approach infinity as x\to \infty . However, as shown in the following table, the values of {x}^{3} are growing much faster than the values of {x}^{2}.

Comparing the Growth Rates of {x}^{2} and {x}^{3}
x 10 100 1000 10,000
f(x)={x}^{2} 100 10,000 1,000,000 100,000,000
g(x)={x}^{3} 1000 1,000,000 1,000,000,000 1,000,000,000,000

In fact, \underset{x\to \infty }{\text{lim}}\ds\,\frac{{x}^{3}}{{x}^{2}}=\underset{x\to \infty }{\text{lim}}x=\infty\ \text{or, equivalently,}\ \underset{x\to \infty }{\text{lim}}\ds\,\frac{{x}^{2}}{{x}^{3}}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{1}{x}=0.

As a result, we say {x}^{3} is growing more rapidly than {x}^{2} as x\to \infty . On the other hand, for f(x)={x}^{2} and g(x)=3{x}^{2}+4x+1, although the values of g(x) are always greater than the values of f(x) for x \symbol{"3E} 0, each value of g(x) is roughly three times the corresponding value of f(x) as x\to \infty , as shown in the following table. In fact,

\underset{x\to \infty }{\text{lim}}\ds\,\frac{{x}^{2}}{3{x}^{2}+4x+1}=\ds\,\frac{1}{3}.
Comparing the Growth Rates of {x}^{2} and 3{x}^{2}+4x+1
x 10 100 1000 10,000
f(x)={x}^{2} 100 10,000 1,000,000 100,000,000
g(x)=3{x}^{2}+4x+1 341 30,401 3,004,001 300,040,001

In this case, we say that {x}^{2} and 3{x}^{2}+4x+1 are growing at the same rate as x\to \infty .

More generally, suppose f and g are two functions that approach infinity as x\to \infty . We say g grows more rapidly than f as x\to \infty if

\underset{x\to \infty }{\text{lim}}\ds\,\frac{g(x)}{f(x)}=\infty\ \text{or, equivalently,}\ \underset{x\to \infty }{\text{lim}}\ds\,\frac{f(x)}{g(x)}=0.

On the other hand, if there exist constants M,m\symbol{"3E}0 and a number a\symbol{"3E}0 such that m\le\ds\,\frac{f(x)}{g(x)}\le M, for all x\ge a, we say that f and g grow at the same rate as x\to \infty.

Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions.

Comparing the Growth Rates of \large{\ds\ln(x),\ {x}^{2},} and  \large{{e}^{x}}

For each of the following pairs of functions, use L’Hôpital’s rule to evaluate \underset{x\to \infty }{\text{lim}}\frac{f(x)}{g(x)}.

  1. f(x)={x}^{2}\text{ and }g(x)={e}^{x}

  2. f(x)=\text{ln}(x)\text{ and }g(x)={x}^{2}

Solution

  1. Since \underset{x\to \infty }{\text{lim}}{x}^{2}=\infty and \underset{x\to \infty }{\text{lim}}{e}^{x}=\infty, we can use L’Hôpital’s rule to evaluate \underset{x\to \infty }{\text{lim}}\ds\,\frac{{x}^{2}}{{e}^{x}}. We obtain \underset{x\to \infty }{\text{lim}}\ds\,\frac{{x}^{2}}{{e}^{x}}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{2x}{{e}^{x}}.

    Because \underset{x\to \infty }{\text{lim}}2x=\infty and \underset{x\to \infty }{\text{lim}}{e}^{x}=\infty , we can apply L’Hôpital’s rule again:

    \underset{x\to \infty }{\text{lim}}\ds\,\frac{2x}{{e}^{x}}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{2}{{e}^{x}}=0.

    Hence, \underset{x\to \infty }{\text{lim}}\ds\,\frac{{x}^{2}}{{e}^{x}}=0, which means that {e}^{x} grows more rapidly than {x}^{2} as x\to \infty (see the figure with the graphs of these functions below).

    The functions g(x) = ex and f(x) = x2 are graphed. It is obvious that g(x) increases much more quickly than f(x).
    Figure 3. An exponential function grows at a faster rate than a power function.
    Growth rates of a power function and an exponential function.
    x 5 10 15 20
    {x}^{2} 25 100 225 400
    {e}^{x} 148 22,026 3,269,017 485,165,195
  2. Since \underset{x\to \infty }{\text{lim}}\text{ln}(x)=\infty and \underset{x\to \infty }{\text{lim}}{x}^{2}=\infty , we can use L’Hôpital’s rule to evaluate \underset{x\to \infty }{\text{lim}}\ds\,\frac{\text{ln}(x)}{{x}^{2}}. We obtain

    \underset{x\to \infty }{\text{lim}}\ds\,\frac{\text{ln}(x)}{{x}^{2}}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{1\text{/}x}{2x}=\underset{x\to \infty }{\text{lim}}\ds\,\frac{1}{2{x}^{2}}=0.

    Thus, {x}^{2} grows more rapidly than \text{ln}(x) as x\to \infty , which agrees with the following figure displaying the graphs of these functions.

    The functions g(x) = x2 and f(x) = ln(x) are graphed. It is obvious that g(x) increases much more quickly than f(x).
    Figure 4. A power function grows at a faster rate than a logarithmic function.
    Growth rates of a power function and a logarithmic function
    x 10 100 1000 10,000
    \text{ln}(x) 2.303 4.605 6.908 9.210
    {x}^{2} 100 10,000 1,000,000 100,000,000

Compare the growth rates of {x}^{100} and {2}^{x}.

Answer

The function {2}^{x} grows faster than {x}^{100}.

Hint

Apply L’Hôpital’s rule sufficiently many times to evaluate \ds\lim_{x\to\infty}\frac{{x}^{100}}{{2}^{x}}

Using the same ideas as in example (a) above, it is not difficult to show that {e}^{x} grows more rapidly than {x}^{p} for any p > 0. In the following table, we compare {e}^{x} with {x}^{3} and {x}^{4} as x becomes large.

An exponential function grows at a faster rate than any power function
x 5 10 15 20
{x}^{3} 125 1000 3375 8000
{x}^{4} 625 10,000 50,625 160,000
{e}^{x} 148 22,026 3,269,017 485,165,195

Similarly, it is also easy to show that {x}^{p} grows more rapidly than \text{ln}(x) for any p > 0. In the table below, we compare \text{ln}(x) with \sqrt[3]{x}=x^{1/3} and \sqrt{x}=x^{1/2} when x becomes large.

A logarithmic function grows at a slower rate than any root function
x 10 100 1000 10,000
\text{ln}(x) 2.303 4.605 6.908 9.210
\sqrt[3]{x} 2.154 4.642 10 21.544
\sqrt{x} 3.162 10 31.623 100

Key Concepts

  • L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form \dfrac{0}{0} or \dfrac{\infty}{\infty} arises.
  • L’Hôpital’s rule can also be applied to other indeterminate forms if the expressions under the limit can be rewritten as a quotient of indeterminate form \dfrac{0}{0} or \dfrac{\infty}{\infty} .
  • The exponential function {e}^{x} grows faster than any power function {x}^{p}, p > 0.
  • The logarithmic function \text{ln}(x) grows slower than any power function {x}^{p}, p > 0.

Exercises

For the following exercises, evaluate the given limit. If the limit does not exist, indicate whether there is a trend of \infty or -\infty. In case you are using L’Hôpital’s rule, explain why it applies.

1. \underset{x\to 3}{\text{lim}}\ds\,\frac{{x}^{2}-9}{x^3+x^2-11x-3}

Answer

\ds\frac3{11}

2. \underset{t\to -1}{\text{lim}}\ds\,\frac{t^7+1}{t^5+1}

3. \underset{x\to 2}{\text{lim}}\ds\,\frac{x^4-5x-6}{x^3-8}

Answer

\dfrac94

4. \underset{t\to \frac{\pi}2}{\text{lim}}\ds\,\frac{\cos(3t)}{ \cot (t)}

5. \underset{x\to 0}{\text{lim}}\ds\,\frac{1-\cos(x)}{x^2}

Answer

\dfrac12

6. \underset{x\to 1}{\text{lim}}\ds\,\frac{x-1}{ \tan (x)}

7. \underset{x\to 0}{\text{lim}}\ds\,\frac{{(1+x)}^{-2}-1}{\sin(x)}

Answer

-2

8. \underset{x\to 0}{\text{lim}}\ds\,\frac{\sqrt{1+x}-\sqrt{1-x}}{x}

9. \underset{x\to \pi }{\text{lim}}\ds\,\frac{x^2-\pi^2 }{ \tan (x)}

Answer

2\pi

10. \underset{t\to \infty}{\text{lim}}\ds\,\frac{\sqrt{t}}{\ln(1+t)}

11. \underset{x\to 0 }{\text{lim}}\ds\,\frac{e^{3x}-1}{ \sin (x)}

Answer

3

12. \underset{x\to 1^-}{\text{lim}}\ds\,\frac{\cos^{-1}(x)}{\sqrt{1-x^2}}

13 \underset{x\to 0}{\text{lim}}\ds\,\frac{3x-7x^5}{\tan^{-1}(2x)}

Answer

\ds\frac32

14. \underset{x\to 0}{\text{lim}}\ds\,\frac{{\pi}^{x}-e^x}{2x^2+x}

15. \underset{t\to \infty}{\text{lim}}\ds\,\frac{\big(\ln(t)\big)^2}{1-t}

Answer

0

16. \underset{x\to a}{\text{lim}}\ds\,\frac{x-a}{{x}^{n}-{a}^{n}}, where a\ne0 is a fixed real number and n is a positive integer.

17. \underset{x\to 0}{\text{lim}}\ds\,\frac{x e^x-x}{\cos(x)-1}

Answer

-2

18. \underset{t\to 0}{\text{lim}}\ds\,\frac{t-\tan(t)}{t^3}

19. \underset{x\to -1}{\text{lim}}\ds\,\frac{x \sin(\pi x)}{3x^2+2x-1}

Answer

\ds-\frac{\pi}4

20. \underset{x\to \infty }{\text{lim}}\ds\,\frac{1-{e}^{x}}{3x+4}.

21. \ds\lim_{x\to-\infty}\,\frac{\ln(1-2x)}{x}

Answer

0

22. \underset{x\to 2}{\text{lim}}\ds\,\frac{ \tan (\pi x)}{\sqrt{2x+5}-3}

23. \underset{x\to 1}{\text{lim}}\ds\,\frac{ \ln (x)}{e^{x^2}-e}

Answer

\ds\frac1{2e}

24. \underset{x\to 0}{\text{lim}}\ds\,\frac{x\cos(x)+\tan(x)}{2x}

25. \underset{x\to 0}{\text{lim}}\ds\,\frac{ 2\sin (x)- \sin (2x)}{{x}^{3}}

Answer

1

26. \underset{x\to 0}{\text{lim}}\ds\,\frac{{e}^{x}-x-1}{{x}^{2}}

27. \underset{x\to \infty}{\text{lim}}\ds\,\frac{x \ln(x)}{1+3x-x^2}

Answer

0

28. \underset{x\to -1}{\text{lim}}\ds\,\frac{x^3+1}{\sec(\pi x)-x}

29. \underset{x\to -1}{\text{lim}}\ds\,\frac{\ln(x^2+x+1)}{x^{100}-x^{10}}

Answer

\ds\frac1{90}

30. \underset{x\to 0^+}{\text{lim}}\ds\,\frac{ \csc(x)}{2^x-1}

31. \underset{x\to 1}{\text{lim}}\ds\,\frac{3^x-\sqrt[3]{x}-2}{1-x}

Answer

\ds\,-3\ln(3)+\frac{1}{3}

32. \underset{x\to {0}^{+}}{\text{lim}}{x}^{2}\,\text{ln}(x)

33. \underset{x\to \infty }{\text{lim}}x \sin \left(\ds\,\frac{1}{x}\right)

Answer

1

34. \ds\lim_{x\to\infty} e^{-\sqrt x}\cdot(3x+1)

35. \ds\underset{x\to \frac{\pi}2}{\text{lim}}\cot(x) \cdot\sec \left(3x\right)

Answer

\ds-\frac13

36. \ds\lim_{x\to0^+}\left(\frac1x-\frac{x+1}{\sin(x)}\right)

37. \ds\lim_{x\to \infty}\Big(\ln(5x^2+3x)-\ln(x^2+1)\Big)

Answer

\ln(5)

38. \underset{x\to 0}{\text{lim}}{(x+1)}^{1\text{/}x}

39. \underset{x\to {1}^{-}}{\text{lim}}{x}^{\csc(1-x)}

Answer

\ds\frac1e

40. \underset{x\to 0^+}{\text{lim}}{\big(\cos(x)\big)^{\sin^{-1}(x)}

41. \underset{x\to 0^+}{\text{lim}}{x}^{\sqrt x}

Answer

1

42. \displaystyle \lim_{x \to 1}\, (4x-3)^{\cot{(\pi x)}}

43. \ds\underset{x\to \infty }{\text{lim}}{\left(1-\frac{1}{x}\right)}^{x}

Answer

\ds\frac{1}{e}

44. \displaystyle \lim_{x \to \infty}\, (2x+5)^{\frac1{1-\ln(x)}}

45. \ds\underset{x\to \infty }{\text{lim}}{\left(x-\ln(x)\right)}

(Hint: Factor x out of the expression.)

Answer

\ds\infty

Glossary

indeterminate forms
when evaluating a limit, the forms \dfrac{0}{0}, \dfrac{\infty}{\infty} , 0\cdot\infty , \infty -\infty , {0}^{0}, {\infty }^{0}, and {1}^{\infty } are considered indeterminate because further analysis is required to determine whether the limit exists and, if so, what its value is
L’Hôpital’s rule
if f and g are differentiable functions over an open interval that contains a, except possibly at a, and \underset{x\to a}{\text{lim}}f(x)=0=\underset{x\to a}{\text{lim}}g(x) or \underset{x\to a}{\text{lim}}f(x) and \underset{x\to a}{\text{lim}}g(x) are infinite, then \ds\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }(x)}{{g}^{\prime }(x)}, provided the limit on the right exists or has a trend of \infty or -\infty

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