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In the application of L'Hospital's rule, we often run into the situation $$\cdots=\lim_{x\to p}\frac{f(x)}{g(x)}=\lim_{x\to p}\frac{f'(x)}{g'(x)}$$

by the quotient rule ( assuming these limits exists ) $$\cdots=\frac{\lim_{x\to p}f'(x)}{\lim_{x\to p}g'(x)}.$$

So we are taking the limit of a derivative at at $p$. However, a derivative is itself a limit by definition. Is the situation then, to be interpreted as such $$\lim_{x\to p} f'(x) = \lim_{(x,h)\to (p,0)}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(p+h)-f(p)}{h}=f'(p)?$$

In most examples, this is resolved by "plugging $p$ to $f'(x)$, so we have $f'(p)$". I think the result follows from the requirement, that we must have differentiability on some interval, where $p\in(a,b)$, but then this would mean L'Hospital doesn't work when we have a function that is differentiable at one point only. What is the correct way to think about $\lim_{x\to p} f'(x)$ when applying L'Hospital's rule?

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    $\begingroup$ A function cannot be differentiable at a point. It is always differentiable in a neighborhood of a point (exactly because it is defined by a limit). $\endgroup$ – Orenio Jun 8 '20 at 15:58
  • $\begingroup$ So if a function is differentiable at $x=0$ only, it is actually differentiable at some neighborhood of $0$? $\endgroup$ – variations Jun 8 '20 at 16:01
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    $\begingroup$ Orenio: What about the function $f(x)$ which is $x^2$ on rationals and $0$ on irrationals? It is discontinuous away from $0$ so definitely not differentiable there. On the other hand, for $h\neq 0$, $0\leq \left| \frac{f(h)}{h}\right|\leq |h|$, so by the squeeze theorem, $f'(0) = 0$. $\endgroup$ – Jason DeVito Jun 8 '20 at 16:04
  • $\begingroup$ "differentiable in a neighborhood" means differentiable in each point of neighborhood. Differentiable in point $p$ of open set needs from function only to be defined in some open set, point belongs to and then existence of well known limit. So function differentiable at point is general definition. $\endgroup$ – zkutch Jun 8 '20 at 16:06
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    $\begingroup$ @Carlo: Perhaps. It is true that if $f'(p)$ exists, then $f$ is continuous at $p$. $\endgroup$ – Jason DeVito Jun 8 '20 at 16:16
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There are two points which you need to understand here.


First you need to observe that one of the hypotheses of L'Hospital's Rule is that the limit of $f'(x) /g'(x) $ exists as $x\to p$. However, this does not necessarily imply that the limits of $f'(x), g'(x) $ exist as $x\to p$. If these limits do exist and limit of $g'(x) $ is non-zero then quotient rule applies as indicated in your question.

But as a thumb rule while applying L'Hospital's Rule always try to think of the ratio $f'(x) /g'(x) $ as a single expression instead of a quotient and try to simplify it and evaluate its limit. The idea behind the rule is that the ratio of derivatives will be simpler in some manner than ratio of the original functions (this is most obvious when we differentiate polynomials and reduce the degree by one).

One must avoid repeated applications of the rule unless absolutely necessary. The typical view of L'Hospital's Rule is differentiate and plug, but in reality it should be differentiate, simplify and evaluate limit.


The second point is more subtle and often overlooked. In order that $\lim_{x\to p}f'(x) $ makes sense a prerequisite is that $f$ must be differentiable in a deleted neighborhood of $p$. If this prerequisite is met then we know that $$f'(x) =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ and thus $$\lim_{x\to p} f'(x) =\lim_{x\to p} \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ The expression on right side in above equation is what one calls an iterated limit and the order of limit operations matters here.

Assuming that we can change the order of limit operation we get $$\lim_{h\to 0}\lim_{x\to p} \frac{f(x+h) - f(x)} {h} $$ Further assuming continuity of $f$ at $p$ we can simplify the expression above as $$\lim_{h\to 0}\frac{f(p+h)-f(p)}{h}=f'(p)$$ provided $f$ is also differentiable at $p$.

So what you write in second part of your question obviously requires differentiability of $f$ at $p$, but more importantly it requires that the order of limit operations can be changed. Unfortunately this is difficult to guarantee.

It is quite possible that $f$ is differentiable at $p$ and yet $\lim_{x\to p} f'(x) $ does not exist. However there is some hope and surprise left due to the following result:

Theorem: If $f$ is continuous at $p$ and $\lim_{x\to p} f'(x) $ exists then $f$ is differentiable at $p$ and $\lim_{x\to p} f'(x) =f'(p) $.

Try to prove the above theorem using definition of derivative and mean value theorem.

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