Worksheet for Section 4.7

Secton 4.7 is about a way of evaluating limits in forms that don't work out well by other techniques, by using derivatives to assist the work of finding the limit. The technique is called L'Hôpital's Rule, and it applies to limits in indeterminate form, such as $0/0$ form or $\infty/\infty$ form. To brush up on limit techniques from earlier chapters, recall that some limits in indeterminate form can be evaluated by applying algebraic techniques:

$$\lim_{x\to 1}\frac{x^2+x-2}{x-1}\qquad \lim_{x\to\infty}\frac{2x-1}{x+1}\qquad \lim_{x\to 0}\frac{e^{2x}-1}{e^x-1}$$

The last example above involves the exponential function, but can still be worked out algebraically. In some cases, however, limits may involve a mixture of transcendental functions and algebraic functions, and in such a case the algebraic techniques may not work. Consider the following limits:

$$\lim_{x\to 0}\frac{e^x-1}{x}\qquad \lim_{x\to 1}\frac{\ln x}{x-1}$$

Both of the above limits are in $0/0$ form -- if you plug directly into the expression, you get 0 in both the numerator and the denominator. But algebraic techniques do not help in evaluating these limits. L'Hôpital's Rule says that if you have an indeterminate form limit, it can be evaluated by taking the derivative of the numerator and the derivative of the denominator:

$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$

Use L'Hôpital's Rule to evaluate the two limits in $0/0$ form above. L'Hôpital's Rule can also be applied to limits in the other indeterminate forms, such as $\infty/\infty$ form, as well as to one-sided limits and limits at infinity. However, sometimes a limit must be rewritten to get it into one of the indeterminate forms to which the Rule applies. Consider, for example, the following limits:

$$\lim_{x\to\infty}\frac{\ln x}{x^2}\qquad \lim_{x\to\infty}\frac{x... ...quad \lim_{x\to\infty} xe^{-x}\qquad \lim_{x\to\infty}\left(1+\frac1x\right)^x$$

The last example above is a particularly interesting one because it has applications to problems involving exponential functions, particularly to compound interest problems. It is in an unusual indeterminate form -- $1^\infty$ form. In order to write this in a form to which L'Hôpital's Rule applies, you will have to write the limit as an equation and take logarithms on both sides:

$$y =\lim_{x\to\infty}\left(1+\frac1x\right)^x$$

$$\ln y =\ln\left[\lim_{x\to\infty}\left(1+\frac1x\right)^x\right]$$

$$\ln y =\lim_{x\to\infty}x\cdot\ln\left(1+\frac1x\right)$$

Now this limit can be written in a form to which L'Hôpital's Rule applies. Here are a couple of other L'Hôpital's Rule problems -- the first one requires a technique similar to the last problem above, and the second one is an example of a one-sided limit to which (with a little bit of rewriting) L'Hôpital's Rule can be applied:

$$\lim_{x\to 0}(1-\cos x)^x\qquad \lim_{x\to 0^+}\left(\frac1{\ln (x+1)}-\frac1{x}\right)$$

Limits at infinity provide a way of comparing the behavior of functions for large values of $x$, and L'Hôpital's Rule can be used again to evaluate the resulting limits. For functions $f(x)$ and $g(x)$, $g(x)$ dominates $f(x)$ as $x\to\infty$ if $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$$. For example, check that $g(x)=\sqrt{x}$ dominates $f(x)=\ln x$ as $x\to\infty$ -- you will need L'Hôpital's Rule to compute the limit.