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I have the following statement to prove:

Prove that if $g$ is derivable, therefore $\lim_{h\to 0}\frac{1}{g(x+h)}=\frac{1}{g(x)}$

My attempt was:

If $g$ is derivable, then is continuous on its domain and the limit of continuous function in a point of its domain is just the image of that point, that is $\frac{1}{g(x)}$.

Is my proof correct?

A second proof that i made was:

I know the fact that $\lim_{h\to 0} g(x+h)=g(x)$ since is continuous.

And i have $[\lim_{h\to 0} g(x+h)]^{-1}=[g(x)]^{-1}$

and since limit exist using limit's algebra, i got the desired result.

Are these proofs correct? thanks in advance.

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    $\begingroup$ you have to assume $g(x) \neq 0$,because otherwise the statement is meaningless. But other than that, yes $\endgroup$
    – peek-a-boo
    Jun 9 '20 at 2:37
  • $\begingroup$ Yes, of course! So, you think that these proof are correct? $\endgroup$ Jun 9 '20 at 2:38
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Of course we assume $g(x) \ne 0$, but nevertheless we need $\lim_{h \rightarrow 0}g(x +h) \ne 0$ in both cases.

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