6
$\begingroup$

Simple question, to which I don't know the answer. Does it work the same even if we are only interested in one-sided limits, and it won't cause problems that the actual limit doesn't exist?

$\endgroup$
3
$\begingroup$

One of the key assumptions of L'Hospital's theorem is that the functions $f,g$ be derivable over an open interval one of whose end points is $c$, the point to which $x$ tends:

$$\lim_{x\rightarrow c}\frac{f(x)}{g(x)}.$$

The proof due to Taylor and presented on Wiki does operate only over this one sided interval of common differentiability.

So the L'Hospital rule is basically about one sided limits.

To be very rigorous, first, one has to examine the derivatives on both sides of $c$. It may turn out that the left and the right limits differ or only one of them exits.

$\endgroup$
0
$\begingroup$

I am not 100% sure what you mean, but let me state the following things and you tell me if that answers your question.

First, let me state l'hopital's rule:

Let $I=(\tilde{x}_0,x_0)$ be a non-empty open Interval and $ f,\, g \colon I\to\Bbb{R} $ differentiable functions, s.t. $x\nearrow x_0$ both functions tend to 0 OR both functions diverge ($-\infty$ or $+\infty$).

If $ g'(x) \neq 0$ $\forall x \in I$ holds and $\tfrac{f'(x)}{g'(x)}$ converges for $x\nearrow x_0$ to $c$ or diverges ($-\infty$ or $+\infty$), then so does $\tfrac{f(x)}{g(x)}$. Analogue for $x\searrow \tilde{x}_0$.

If $I$ is fully contained in an open intervall, where the above conditions hold, then especially the following holds$$\lim_{x\to x_0}\frac{f'(x)}{g'(x)}=c~\Rightarrow~\lim_{x\to x_0}\frac{f(x)}{g(x)}=c.$$ This also holds true for interval boundaries $x_0=\pm\infty$.

Secondly, recall that:

By definition the limit $\lim_{x\rightarrow x_0}f(x)$ exists if and only if $\lim_{x\searrow x_0}f(x)$ and $\lim_{x\nearrow x_0}f(x)$ exist.

Hence, for the statement you only need onesided limits, and the existence of the bothsided limit in the result implies the existence of the onesided ones.

$\endgroup$

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.