Change of Variables in Limits (Part 1)

You certainly need some assumptions on $f,g$, as Jared shows.

THM Suppose that $f(y)\to \ell$ as $ y\to b$. Suppose that ${\rm im}\, g\subseteq {\rm dom}\, f$, and suppose that $g(x)\to b$ as $x\to a$, yet $g$ does not attain the value $b$ in a neighborhood $B(x,\eta)-\{x\}$. Then $$f\circ g(x)\to \ell \;\;\text{ as } x\to a$$

P Let $\epsilon >0$ be given. Since $f\to\ell $ as $y\to b$ there exists $\delta >0$ such that $0|y-b|<\delta$ implies $|f(y)-\ell|<\epsilon$. Since $g\to b$ as $x\to a$, there exists $\eta >\delta'>0$ such that $0<|x-a|<\delta'$ implies $0<|g(x)-b|<\delta$. But then we will have $|f(g(x))- \ell|<\epsilon$ whenever $0<|x-a|<\delta'$, so the claim follows. $\blacktriangle$

This then gives the standard

COR Suppose that $f$ is continuous at $g(a)$ and $g$ is continuous at $a$. Then $f\circ g$ is continuous at $a$.

A counterexample. Let $a=b=0$, $g(x)=x^2$, and let

$$f(x)=\begin{cases}1&x\ge 0\\0&x<0\end{cases}$$

Then $\lim_{y\to 0}f(y)$ does not exist, but $\lim_{x\to 0}f(g(x))=1$.

If we suppose that $\lim_{y\to b}f(y)$ exists, then the result holds.

First suppose that the limit is finite and equal to $L$. Fix $\epsilon>0$, and let $\delta$ be such that

$$|y-b|<\delta\Longrightarrow |f(y)-L|<\epsilon$$

Now, choose $\delta'$ such that

$$|x-a|<\delta'\Longrightarrow|g(x)-b|<\delta$$

Notice now that we have

$$|x-a|<\delta'\Longrightarrow|f(g(x))-L|<\epsilon$$

which shows that $\lim_{x\to a}f(g(x))=L$. The case of an infinite limit is shown similarly.