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Let $f : \mathbb{R} \to \mathbb{R}$ be continous. Assume that $f’(x)$ exists for all $x \neq 0$ and $ \lim_{x\to\ 0} f'(x) = 1$. Show that $f’(0)$ exists and $f’(0) = 1$

My attempt: $$1 = \lim_{x\to0} \lim_{h\to0}\frac{f(x + h) - f(x)}{h} = \lim_{h\to0}\frac{f(0 + h) - f(0)}{h} = f’(0)$$

I don’t think that the limit interchange that I have done is correct. Can someone help me out with how to do this.

gtoques
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user601297
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1 Answers1

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I think the post Martin R linked says something similar, but this is a standard application of the MVT: Fix $h>0$ and consider $\frac{f(h) - f(0)}{h}$, then by the mean value theorem you can find a point $a \in (0,h)$ such that $\frac{f(h) - f(0)}{h} = f'(a)$. Now take $h \to 0$. What happens to $a$? Keep in mind that $a$ depends on $h$.

Also, interchanging limits is not a good idea unless you're appealing to a specific theorem/result that lets you do this. In general even "easy" limits can't be changed.

gtoques
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  • When you ask me what happens to $a$ ( continuity if $f’$) are you suggesting that $f’$ is continous? Because that’s not given in the question. – user601297 Dec 12 '20 at 08:23
  • Oh, my bad, I didn't read that. You don't need continuity. I've edited my answer -- it's almost the same. The idea is that as $h \to 0$, $a \to 0$ so the RHS goes to $1$ and hence the LHS goes to $1$ (and exists). – gtoques Dec 12 '20 at 08:26
  • Yes i got it, thanks a lot – user601297 Dec 12 '20 at 08:31