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

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This looks like homework; please read…;. –  Nate Eldredge Apr 11 '12 at 2:30
Yes, it is. Sorry about that. New to the site, thanks for the heads up. –  JackReacher Apr 12 '12 at 1:39

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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.

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

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