How do you prove the property that you can move limits in and out of continuous functions?

Question

Suppose $$\lim_{x\to c} f(x) = L,$$ and if $g$ is continuous, then $$g\left(\lim_{x \to c} f(x)\right) = g(L) \Rightarrow \lim_{x\to c} g(f(x)) = g(L).$$ A popular application of this would be to take the logarithm of both sides and then move the logarithm inside the limit. How do you prove this property?

Answer 1

It's straight from definition of continuous and limits.

Let $\lim_{x\to c}f(x) = L$. $g$ is continuous so for any $\epsilon > 0$ there is a $\delta > 0$ so that $|y-L| < \delta \implies |g(y) - g(L)| < \epsilon$.

But $\lim_{x\to c}f(x) = L$ so for that $\delta$ then there is $\gamma > 0$ so that if $|x - c|< \gamma$ then $|f(x) - L| < \delta$ and therefore $|g(f(x)) - g(L)| = |g(f(x)) - g(\lim_{x\to c}f(x))| < \epsilon$.

So by definition. $\lim_{x\to c} g(f(x)) = g(L)=g(\lim_{x\to c} f(x))$.

Answer 2

Let $\epsilon>0$. By continuity at $L$, there exists $\eta>0$ such that $|f(x)-L|<\eta \implies |g(f(x))-g(L)\rvert<\epsilon$. Since $\lim_{x\to c} f(x)=L$, there exists $\delta>0$ such that $0<|x-c|<\delta \implies |f(x)-L|<\eta$. Hence, $0<|x-c|<\delta \implies |g(f(x))-g(L)|<\epsilon$.

By definition, $$ \lim_{x\to c} g(f(x))=g\left(\lim_{x\to c}f(x)\right)=g(L). $$