Suppose $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$. Let $\epsilon$ be given. Then, there exists a $\delta_1$ such that $|h|<\delta_1 \implies \left|\frac{g(f(a)+h)-g(f(a))}{h} - g'(f(a))\right| <\epsilon$. Since $\lim \limits_{h \to 0} f(a+h) -f(a) = 0$, there is a $\delta_2$ s.t. $|h| < \delta_2 \implies |f(a+h)-f(a)|<\delta_1$.
Suppose $|h| < \delta_2$. Then, $f(a)+f(a+h)-f(a) = f(a+h)$, so $\left| \frac{g(f(a+h))-g(f(a))}{f(a+h)-f(a)} - g'(f(a))\right| < \epsilon$, and so $\lim \limits_{h \to 0} \frac{g(f(a+h))-g(f(a))}{f(a+h)-f(a)} = g'(f(a))$. Since $\lim \limits…
There are two definitions on the limit of a function:
Definition 1: Let f(x) be a function defined on a neighbourhood of $x_0$, $\lim_{x\to x_0} f(x)=A $ if only if for every real $\varepsilon>0$, there exists a real $ \delta>0$ such that for all $\mathrm{x}, 0<|x-p|<\delta$ implies that ...
