# Thorem: If f(x) is continuous at L and lim_(x->a) g(x)=L, then lim_(x->a) f(g(x)=f(lim_(x->a) g(x))=f(L). Proof: Assume f(x) is continuous at a point L, and that lim_(x->a) g(x)=L.

Thorem: If $f\left(x\right)$ is continuous at $L$ and $\underset{x\to a}{lim}g\left(x\right)=L$, then $\underset{x\to a}{lim}f\left(g\left(x\right)=f\left(\underset{x\to a}{lim}g\left(x\right)\right)=f\left(L\right)$.
Proof: Assume $f\left(x\right)$ is continuous at a point $L$, and that $\underset{x\to a}{lim}g\left(x\right)=L$.
$\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:\left[|x-L|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(x\right)-f\left(L\right)|<\epsilon \right]$.
And $\mathrm{\forall }\delta >0,\mathrm{\exists }{\delta }^{\prime }>0:\left[|x-a|<{\delta }^{\prime }\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|g\left(x\right)-L<\delta \right]$.
So, $\mathrm{\forall }\delta >0,\mathrm{\exists }{\delta }^{\prime }>0:\left[|x-a|<{\delta }^{\prime }\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(g\left(x\right)\right)-f\left(L\right)|<ϵ\right]$.
$\underset{x\to a}{lim}g\left(x\right)=L$ so $f\left(\underset{x\to a}{lim}g\left(x\right)\right)=f\left(L\right)$.
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wegpluktee3
You're doing the right thing but the way you've presented it is a bit confusing. Why not word it like this:
Pick $ϵ>0$. Continuity of $f$ at $L$ gives you an $\eta$ such that $|x-L|<\eta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(x\right)-f\left(L\right)|<ϵ$.
For this $\eta$, $\underset{x\to a}{lim}g\left(x\right)=L$ gives you a $\delta$ such that $|x-a|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|g\left(x\right)-L|<\eta$.
Hence $|x-a|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|g\left(x\right)-L|<\eta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(g\left(x\right)\right)-f\left(L\right)|<ϵ$