3.6. Generalizations of L'Hôpital's RuleMathematical theorems are sometimes like cars. I own a Honda Fit that is about as barebones as you can get these days, but persuading a dealer to sell me that car was like pulling teeth. The salesman was absolutely certain that any sane customer would want to pay an extra $1,800 for such crucial amenities as floor mats and a chrome tailpipe. L'Hôpital's rule in its most general form is a much fancier piece of machinery than the stripped down model described on p. 61. The price you pay for the deluxe model is that the proof becomes much more complicated than the oneliner that sufficed for the simple version. Multiple Applications of the RuleIn the following example, we have to use l'Hôpital's rule twice before we get an answer. Example 47◊ Evaluate \[ \lim_{x\rightarrow \pi} \frac{1+\cos x}{(x\pi)^2} \] ◊ Applying l'Hôpital's rule gives \[ \frac{\sin x}{2(x\pi)} , \] which still produces 0/0 when we plug in x=π. Going again, we get \[ \frac{\cos x}{2} = \frac{1}{2} . \] The reason that this always works is outlined on p. 152. The Indeterminate Form ∞/∞Consider an example like this: \[\lim_{x\rightarrow 0} \frac{1+1/x}{1+2/x} .\] This is an indeterminate form like ∞/∞ rather than the 0/0 form for which we've already proved l'Hôpital's rule. As proved on p. 153, l'Hôpital's rule applies to examples like this as well. Example 48◊ Evaluate \[ \lim_{x\rightarrow 0} \frac{1+1/x}{1+2/x} . \] ◊ Both the numerator and the denominator go to infinity. Differentiation of the top and bottom gives (x^{2})/(2x^{2}) = 1/2. We can see that the reason the rule worked was that (1) the constant terms were irrelevant because they become negligible as the 1/x terms blow up; and (2) differentiating the blowingup 1/x terms makes them into the same x^{2} on top and bottom, which cancel. Note that we could also have gotten this result without l'Hôpital's rule, simply by multiplying both the top and the bottom of the original expression by x in order to rewrite it as (x+1)/(x+2). Limits at InfinityIt is straightforward to prove a variant of l'Hôpital's rule that allows us to do limits at infinity. The general proof is left as an exercise (problem 8, p. 68). The result is that l'Hôpital's rule is equally valid when the limit is at ±∞ rather than at some real number a. Example 49◊ Evaluate \[ \lim_{x\rightarrow \infty} \frac{2x+7}{x+8686} . \] ◊ We could use a change of variable to make this into example 39 on p. 59, which was solved using an ad hoc and multiplestep procedure. But having established the more general form of l'Hôpital's rule, we can do it in one step. Differentiation of the top and bottom produces \[ \lim_{x\rightarrow \infty} \frac{2x+7}{x+8686} = \frac{2}{1} = 1 . \]
