Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Specifically, if $\lim\limits_{x \to a} g(x) = b$, and if $f$ is continuous at$ b$, then the $\lim\limits_{x \to a} f \circ g(x) = f(b)$.

share|improve this question
    
This looks like homework; please read meta.math.stackexchange.com/questions/1803/…;. –  Nate Eldredge Apr 11 '12 at 2:30
1  
Yes, it is. Sorry about that. New to the site, thanks for the heads up. –  JackReacher Apr 12 '12 at 1:39

1 Answer 1

up vote 1 down vote accepted

We need to show given $\epsilon > 0$, there exists a $\delta > 0$ so that $|x-a| < \delta \Rightarrow |f(g(x))-f(b)|<\epsilon$. So first find $\delta' > 0$ so that $|z-b| < \delta' \Rightarrow |f(z)-f(b)| < \epsilon$

(we can find a $\delta'$ because $f$ is continuous at $b$).

Then find $\delta > 0$ so that $|x-a| < \delta \Rightarrow |g(x)-g(a)| < \delta'$

(we can find a $\delta$, because $g$ is continuous at $a$).

Now collecting this together, if $|x-a| < \delta$ then $|g(x)-g(a)|=|g(x)-b| < \delta'$. But ("taking" $z=g(x)$), this means that $|f(g(x))-f(b)| < \epsilon$, as desired.

share|improve this answer
    
Thank you John. So clear and concise. Very helpful. –  JackReacher Apr 11 '12 at 9:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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