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.

  • 2
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.