# Using L'Hospital's rule to prove the following

$$f'(x)= \lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}.$$ It is given that the derivative of $f$ at $x$ exists.

If I use L'Hospital's rule and differentiate top and bottom wrt $h$, I get

$$\lim_{h \to 0} \frac {f'(x+h) + f'(x-h)} {2}$$

Then what? I am really lost on this one. I can get the result without using the rule, but the challenge is to use it. Any ideas?

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The stated equality can be proved, without extra assumptions, working with the definition of differentiability. In order to solve it using per se L'Hopitals one must make the extra (stronger) assumption that f' is continuous at x. – Ittay Weiss Jan 11 '13 at 3:11

The question is (implicitly) assuming that the derivative of $f$ is continuous at $x$. Note that you forgot the inner derivative of the second term in the numerator, so that the limit you got is not correct. The correct limit you get is $\lim_{h\to 0} (f'(x+h)+f'(x-h))/2$ which, by continuity of $f'$ at $x$, evaluates to $(f'(x)+f'(x))/2=f'(x)$.

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Thank you so much for explaining this. But what I fail to see is how, and I would be really grateful if you can kindly take your time to explain this to me, $f'(x+h) + f'(x-h) = 2f'(x)$. More specifically, why is it that $f'(x+h) = f'(x)$ or $f'(x-h)= f'(x)$? I would really appreciate any help. – user43901 Jan 11 '13 at 2:59
Or is it that we are taking the limit as $h \rightarrow 0$ which is why we are getting $f'(x+h) = f'(x)$? How does the continuity of $f'$ come into play here? – user43901 Jan 11 '13 at 3:00
@IttayWeiss I think the question doesn't assume that. If one tries to use L'H to solve it, one needs that extra assumption, but the question can be solved without the extra assumption other ways... – N. S. Jan 11 '13 at 3:00
none of the equalities you mention are true. What is true is that under the extra assumption of continuity of f' at x the limit of f'(x+h), as h goes to 0, is f'(x). Similarly for the limit of f'(x-h). Without the extra assumption of continuity of f' at x you can't use L'Hopitals to prove the result (though the result remains true). – Ittay Weiss Jan 11 '13 at 3:01
@ N.S. I don't know exactly how the original question is formulated. If it requires to use L'Hopitals then it must make the extra assumption of continuity of f' at x, otherwise that approach won't work. Of course, this is a very strange question since, of course, the claim is true always. – Ittay Weiss Jan 11 '13 at 3:03

L'Hospital is the wrong approach since you only know that $f$ is differentiable at $x$, and you don't know that $f'$ is continuous. That is what you need for that approach.

Anyhow

$$\lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}=\frac{1}{2} \left[ \lim_{h \to 0} \frac {f(x+h)- f(x)}{h}+ \lim_{h \to 0} \frac {f(x)- f(x-h)}{h}\right]$$

By the definition of differentiability at $x$ we have

$$\lim_{h \to 0} \frac {f(x+h)- f(x)}{h}=f'(x)$$

and after the substitution $h=-t$ you get

$$\lim_{h \to 0} \frac {f(x)- f(x-h)}{h}=\lim_{t \to 0} \frac {f(x)- f(x+t)}{-t} =\lim_{t \to 0} \frac {f(x+t)- f(x)}{t} =f'(x)$$

thus

$$\lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}=\frac{1}{2} [ f'(x)+f'(x)]=f'(x) \,.$$

P.S. If the problem asks you explicitly to use L'H, and you don't have the extra condition that $f$ is differentiable around $x$ and $f'$ is continuous at $x$, then the problem is wrong.

Let $f(x)=x^2\sin(\frac{1}{x})$ with $f(0)=0$. Then by the Squeze Theorem you can prove that $f'(0)=0$.

But

$$f'(x)=2x \sin(\frac{1}{x})-\cos(\frac{1}{x})$$

and hence

$$\lim_{h \to 0} \frac {f'(0+h)+ f'(0-h)}{2}=\lim_{h \to 0} \frac {2h \sin(\frac{1}{h})-\cos(\frac{1}{h})+2h \sin(\frac{1}{h})-\cos(\frac{1}{h})}{2}=\mbox{Does Not Exist}$$

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OP explicitly indicated an interest in a proof using L'Hopitals. – Ittay Weiss Jan 11 '13 at 3:13
The problem does indicate the derivability of $f$ at $x$; however, it does not state the extra assumption. Let me look into it again. – user43901 Jan 11 '13 at 3:17
@IttayWeiss I saw but that is odd since any proof involving L'H needs extra assumptions, which are actually not needed... – N. S. Jan 11 '13 at 3:18
@user43901 Without the extra assumption, use the counterexample I provided with $x=0$ ;) – N. S. Jan 11 '13 at 3:19
I completely agree with you guys! Another question, how can one go about visualizing this process? Can there be a picture one can draw to substantiate the proof IF the extra assumption is given? – user43901 Jan 11 '13 at 3:21

$$f'(x)= \lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}=\lim_{h \to 0} \frac {f'(x+h) + f'(x-h)} {2} = \frac{2f'(x)}{2}=f'(x)$$

though it's a bit circular.

Your mistake was when you differentiated, you forgot to change the sign before $f'(x-h)$.

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Why is $\lim_{h \to 0} f'(x+h)=f'(x)$? And why does $f'(x+h)$ even exist? – N. S. Jan 11 '13 at 2:50
There is nothing that prevents $h$ from being zero like before (where a 0/0 results)\, and because of continuity, I can simply let $h=0$. – Argon Jan 11 '13 at 2:51
The problem only tells you that $f$ is differentiable at $x$. And the derivative of a function is not necessarily continuous.... Try taht with $f(x)=x^2 \sin(\frac{1}{x})$. – N. S. Jan 11 '13 at 2:52
@N.S. If it is not continuous there, then it is not differentiable there, no? – Argon Jan 11 '13 at 2:53
@Aragon Yes you are right if $f$ is differentiable then $f$ is continuous. BUT you are using that $f'$ is continuous, not that $f$ is. And that is not true... – N. S. Jan 11 '13 at 2:58