l'Hôpital revisited

We can get a much more useful form of l'Hôpital's rule using the Cauchy Mean Value Theorem, rather than working, as we did in 5.9, directly from the definition of the derivative.

Proposition 5.22 (l'Hôpital's rule: general form)   . Let f and g be functions such that f (a) = g(a) = 0, and suppose that f and g are differentiable on an open interval I containing a, and that g'(x)$\ne$ 0, except perhaps at a. Then

$$\lim_{{x \to a}}^{}$$$${\frac{{f(x)}}{{g(x)}}}$$ = $$\lim_{{x \to a}}^{}$$$${\frac{{f'(x)}}{{g'(x)}}}$$,

provided the second limit exists.

Proof. Pick x > a and apply the Cauchy Mean Value Theorem to the interval [a, x], to find c with a < c < x such that

$${\frac{{f(x)}}{{g(x)}}}$$ = $${\frac{{f(x) - f(a)}}{{g(x) - g(a)}}}$$ = $${\frac{{f'(c)}}{{g'(c)}}}$$.

Then $$\lim_{{x \to a+}}^{}$$$${\frac{{f(x)}}{{g(x)}}}$$=$$\lim_{{c\to a+}}^{}$$$${\frac{{f'(c)}}{{g'(c)}}}$$=$$\lim_{{x \to a+}}^{}$$$${\frac{{f'(x)}}{{g'(x)}}}$$, since we know the actual limit (not just the one sided limit) exists. Now repeat with x < a to get the result. $\qedsymbol$

Example 5.23   Evaluate $$\lim_{{x \to 0}}^{}$$$${\frac{{1 - \cos x}}{{x^2}}}$$=$${\frac{{1}}{{2}}}$$.

Solution. We have

$$\lim_{{x \to 0}}^{}$$$${\frac{{1 - \cos x}}{{x^2}}}$$ = $$\lim_{{x \to 0}}^{}$$$${\frac{{\sin x}}{{2x}}}$$ = $${\frac{{1}}{{2}}}$$,

where the use of l'Hôpital is justified since the second limit exists. Note that you can't differentiate top and bottom again, and still expect to get the correct answer; one of the hypotheses of l'Hôpital is that the first quotient is of the 0/0 form.

Example 5.24   Use l'Hôpital to establish the following:

$$\lim_{{x \to 0}}^{}$$$${\frac{{\sqrt{1+x}-1-x/2}}{{x^2}}}$$ = - $${\frac{{1}}{{8}}}$$.

Solution. We have

$$\lim_{{x \to 0}}^{}$$$${\frac{{\sqrt{1+x}-1-x/2}}{{x^2}}}$$ = $$\lim_{{x \to 0}}^{}$$$${\frac{{(1/2)(1+x)^{-1/2}-1/2}}{{2x}}}$$ = $$\lim_{{x \to 0}}^{}$$$${\frac{{(1/2)(-1/2)(1+x)^{-3/2}}}{{2}}}$$ = $${\frac{{-1}}{{8}}}$$,

The use of l'Hôpital is justified the second time, since the third limit exists; since we now know the second limit exists, the use of l'Hôpital is justified the first time.

Exercise 5.25   Evaluate $$\lim_{{x \to 0}}^{}$$$$\left(\vphantom{\frac{1}{\sin x} - \frac{1}{x}}\right.$$$${\frac{{1}}{{\sin x}}}$$ - $${\frac{{1}}{{x}}}$$$$\left.\vphantom{\frac{1}{\sin x} - \frac{1}{x}}\right)$$=$$\lim_{{x \to 0}}^{}$$$${\frac{{x - \sin x}}{{x\sin x}}}$$.