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Suppose $f$ is differentiable on $[0,\infty)$. One may write $$ \frac{f(t)-f(0)}{t-0} = f'(\epsilon) $$ for $t\in (0,1),\epsilon \in (0,t)$ by the Mean Value Theorem. I would like to then take the limit as $t\to 0$ (so $\epsilon \to 0$ as well) and say that this equals $f'(0)$. But I think this would require continuity of the derivative, because we don't know how $\epsilon$ changes.

However, if we just apply the definition of $f$ being differentiable we get $$ \lim_{t\to 0}\frac{f(t)-f(0)}{t-0} = f'(0) $$ by definition.

This doesn't feel right to me. I wouldn't think that applying the Mean Value Theorem would lose so much information that I cannot justify the limit anymore.

I there a way to show the limit exists using the Mean Value Theorem like I tried? (Of course using the definition is better, but I just want to know if there is a way to do it with MVT).

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I think that in this case f'(epsilon) is a composition of functions that tends to f'(0).The limit of the derivate can exist or not, if it does exist it is equal to the value at zero. – alpha.Debi Feb 5 '12 at 15:40
I'm a little confused about what you'd like to do, formally, here. You want to show that the limit exists using MVT. However, you already know the limit exists. Perhaps you'd like to remove first the hypothesis $f$ is differentiable at $0$. Am I making sense? – student Feb 5 '12 at 15:41
up vote 3 down vote accepted

I'm not sure if this is what you're asking about, but the limit $\lim\limits_{\epsilon\rightarrow0} f'(\epsilon)$ need not exist.

Consider $f(x)=\cases{ x^2\cos(1/x), & $x\ne0$\cr 0,&$x=0$}$.


$$f'(0)=\lim\limits_{h\rightarrow0}{ h^2\cos(1/x)\over h}=0.$$

While for $x\ne0$ $$ f'(x)=2x\cos(1/x)+\sin(1/x). $$

From the above, it is seen that $f'$ takes values less than $-1/2$ and greater than $1/2$ in any nhood of $0$; thus $\lim\limits_{\epsilon\rightarrow0} f'(\epsilon)$ does not exist.

However, if $\lim\limits_{\epsilon\rightarrow0} f'(\epsilon)$ exists, one can use the MVT to show it is indeed $f'(0)$ (and in fact one need only assume $f'$ exists on $(0,\infty)$, here).

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I think what he said in your example shows that the composition of the derivate with the epsilons of the TVM of each t has limit f'(0), but the function f' not necesary.. – alpha.Debi Feb 5 '12 at 15:56

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