understanding a step in the proof of the chain rule statement of the chain rule:
let $f: A \to R$ and $g: B \to R$. s.t. $f(A) \subset B$ where $ g ∘ f$ is defined. If $f$ is differentiable at $a \in A$ and $g$ is  differentiable at $f(a) \in B $, then  $ g ∘ f$ is differentiable at $a$ with $ (g ∘ f)' (a)=g'(f(a))f'(a)$.
this is the proof that was given:

*

*$g'(f(a))$ exists by assumption so: $g'(f(a))=$$\lim_{y \to f(a)} \frac{g(y)-g(f(a))}{y-f(a)}$


*$f'(a)$ exists by assumption so: $f'(a)=$ $\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$


*$f$ is differentiable at $a$ $\implies $ $f$ is continuous at $a$ $\implies$ $\lim_{x \to a} f(x)=f(a)$


*step 3 implies: $g'(f(a))= \lim_{x \to a} \frac{g(f(x))-g(f(a))}{f(x)-f(a)}$


*so, multiplying two limits together we get : $g'(f(a))f'(a) = \lim_{x \to a} \frac{g(f(x))-g(f(a))}{f(x)-f(a)}\frac{f(x)-f(a)}{x-a}=\lim_{x \to a} \frac{g(f(x))-g(f(a))}{x-a}= (g ∘ f)' (a)$
I do not understand step 4. in the proof. how does step 3 imply this equality of the limit in the RHS of 1 with the limit in the RHS of 4.? Can you please, as simply and with as much detail as possible help me understand?
 A: It is easiest to see this using the sequential criterion for limits.
In Step 1: $g'(f(a)) = \lim_{y \to f(a)} \frac{g(y) - g(f(a))}{y - f(a)}$ can be rephrased as: for all sequences $\{y_n\}_{n = 1}^\infty$ converging to $f(a)$ we must have $g'(f(a)) = \lim_{n \to \infty} \frac{g(y_n) - g(f(a))}{y_n - f(a)}$.
In Step 3: $\lim_{x \to a} f(x) = f(a)$ can be rephrased as: for all sequences $\{x_n\}_{n = 1}^\infty$ converging to $a$ we must have $\lim_{n \to \infty} f(x_n) = f(a)$.
Now let us finally look at your question:
In Step 4: we need to show $g'(f(a)) = \lim_{x \to a} \frac{g(f(x)) - g(f(a))}{f(x) - f(a)}$. To do this with the sequential criterion, we need to consider some arbitrary sequence $\{x_n\}_{n = 1}^\infty$ converging to $a$.
But if such a sequence $x_n \to a$ is given, notice that $\{f(x_n)\}_{n = 1}^\infty$ is a new sequence converging to $f(a)$ by our rephrasing $\lim_{n \to \infty} f(x_n) = f(a)$ of Step 3 above.
But then, by our rephrasing $g'(f(a)) = \lim_{n \to \infty} \frac{g(y_n) - g(f(a))}{y_n - f(a)}$ of Step 1 above, we can set $y_n := f(x_n)$ to get $g'(f(a)) = \lim_{n \to \infty} \frac{g(y_n) - g(f(a))}{y_n - f(a)} = \lim_{n \to \infty} \frac{g(f(x_n)) - g(f(a))}{f(x_n) - f(a)}$.
And this is exactly what it means to have $g'(f(a)) = \lim_{x \to a} \frac{g(f(x)) - g(f(a))}{f(x) - f(a)}$ according to the sequential criterion.
