If you have an indeterminate form:
- $\frac{-\infty}\infty$
- $\frac\infty{-\infty}$
- $\frac{-\infty}{-\infty}$
does l'Hôpital's rule also apply?
$\endgroup$3 Answers
$\begingroup$Yes, but you should first make sure whether or not you really need L'Hopital's rule! If you apply it to stuff which doesn't require you to, weird things will happen.
Take for example the following limit:
$\endgroup$ $\begingroup$$\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x}$.
Blindly applying L'Hopital's Rule repeatedly gives: $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x} = \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+}\dfrac{-1/x^2}{2/x^3} = \lim_{x \to 0^+}\dfrac{2/x^3}{-6/x^4} = \cdots$.
But it we stop after applying L'Hopital's Rule once and simplify stuff, we get: $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x} = \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$.
L'Hopital can be used in all cases of $\infty$ in the denominator; it doesn't matter what the numerator is doing. (Compare with Stolz-Cesaro.)
$\endgroup$ $\begingroup$Yes, it does apply,
Consider taking limits of these, by using l.Hopital
$1$) $\lim_ {x- \to \infty} \dfrac{-x}{x}$
$2$) $\lim_ {x- \to \infty} \dfrac{x}{-x}$
$3$) $\lim_ {x- \to \infty} \dfrac{-x}{-x}$
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