# How to prove that the limit is equal to the function?

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.

• you have to assume $g(x) \neq 0$,because otherwise the statement is meaningless. But other than that, yes Jun 9 '20 at 2:37
• Yes, of course! So, you think that these proof are correct? Jun 9 '20 at 2:38

Of course we assume $$g(x) \ne 0$$, but nevertheless we need $$\lim_{h \rightarrow 0}g(x +h) \ne 0$$ in both cases.