2
$\begingroup$

In a source I have been reading, this statement was made regarding L'Hôpital's rule:

Blockquote

Why is it the case that L'Hôpital's rule is applicable only if the right-hand limit exists? Why not the left-hand? Why not both? I have read other sources on L'Hôpital's rule that do not mention it and would like clarification.

Here is the text:

L'Hôpital's Rule: Suppose that $f$ and $g$ are differentiable functions, and $f(a)=g(a)=0$, and suppose that $g'(x)$ is nonzero in a neighborhood of $a$ (except maybe at $a$ itself). Then $$\lim\limits_{x \to a} \ \frac{f(x)}{g(x)}=\lim\limits_{x \to a} \ \frac{f'(x)}{g'(x)}$$ if the limit on the right-hand side exists.

$\endgroup$
  • $\begingroup$ I don't think the book gives enough clarification. L'hopital can still be useful for limits that don't exist in a rigourous sense. An example would be the limit $lim_{x\rightarrow \infty} (x^2/x)$ which tends to infinity. The limit doesn't exist, but L'hopital is still useful. $\endgroup$ – Hongyu Wang Oct 3 '18 at 21:05
  • $\begingroup$ "Why is it the case that L'Hôpital's rule is applicable only if the right-hand limit exists?" How did you get that idea? $\endgroup$ – zhw. Oct 3 '18 at 21:07
  • $\begingroup$ @zhw My reasoning is that it says that L'Hôpital's rule is true if specifically the right-hand limit exists; I took this to mean that for whatever reason the right-hand limit in particular must exist if L'Hôpital's rule is to hold. If it does not, then as it the text says, L'Hôpital's rule cannot (or, at least, should not) apply. $\endgroup$ – Descartes Before the Horse Oct 3 '18 at 21:22
  • $\begingroup$ If you could write down exactly what the book says, I think we can straighten this out. $\endgroup$ – zhw. Oct 3 '18 at 21:28
  • $\begingroup$ The left limit might exist even though the functions $f,g$ are not differentiable. E.g., $|x|/|x|$. $\endgroup$ – Yves Daoust Oct 3 '18 at 22:39
3
$\begingroup$

There are cases where $\lim\frac{f(x)}{g(x)}$ exists but $\lim\frac{f'(x)}{g'(x)}$ does not exist. For example, with $a=0$:

$$ f(x) = x^2\sin(1/x) \qquad\qquad g(x) = x $$

Here $\lim\limits_{x\to 0}\frac{f(x)}{g(x)}=0$, but $\frac{f'(x)}{g'(x)}$ does not have a limit for $x\to 0$.

Therefore L'Hospital's rule can only go in one direction: If ${f'(x)}/{g'(x)}$ happens to have a limit (and $f(x), g(x)$ both tend to $0$ or $\infty$), then this is also the limit of ${f(x)}/{g(x)}$.

But if ${f'(x)}/{g'(x)}$ does not exist, then this is not enough to conclude anything about whether ${f(x)}/{g(x)}$ has a limit or not.

$\endgroup$
  • $\begingroup$ Just to add here, the existence of limit of $f'/g'$ is to include finite as well as infinite limit also so that if $f'/g'\to\infty$ then $f/g\to\infty$. $\endgroup$ – Paramanand Singh Oct 4 '18 at 6:48
2
$\begingroup$

My pocket example of such a limit is $$ \lim_{x \to \infty} \frac{x + \sin x}{x + \cos x}.$$ This limit is very clearly $1$. But an application of l'Hopital's rule would lead to the consideration of $$ \lim_{x \to \infty} \frac{1 + \cos x}{1 - \sin x},$$ which doesn't exist! Thus existence of the left hand limit does not guarantee the existence of the right hand limit.

$\endgroup$
  • 1
    $\begingroup$ Note that this example does not satisfy the assumption that $g'(x)\ne 0$ in a neighborhood of $a$. (But there are very slight variations of it that do). $\endgroup$ – hmakholm left over Monica Oct 3 '18 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.