Theorem: If $f$ is differentiable at $a$, then $f$ is continuous at $a$.
Proof: This is just a matter of recalling the definitions of differentiability and continuity, and then using the clever trick of subtracting and adding $f(a)$ and multiplying and dividing by $x\ -\ a$.

Assume $f$ is differentiable at $a$.

This means $f\ '(a)=\lim_{x\to a}\ \frac{(f(x)\ -\ f(a))}{(x\ -\ a)}$ exists.

We want to show that $f$ is continuous at $a$, i.e.

$\lim_{x\to a}\ f(x)=f(a)$.

So consider $\lim_{x\to a}\ f(x)=\lim_{x\to a}\ (f(x)\ -\ f(a)+f(a))$ (clever trick 1)

$=\lim_{x\to a}\ (\frac{(f(x)\ -\ f(a))}{(x\ -\ a)})(x\ -\ a)+f(a))$ (clever trick 2)

$=\lim_{x\to a}\ \frac{(f(x)\ -\ f(a))}{(x\ -\ a)}\lim_{x\to a}\ (x\ -\ a)+ \lim_{x\to a}\ f(a))$ using the sum and product law for limits.

$=f\ '(a)0+f(a)=f(a)$. This proves that $f$ is continuous at $a$.

Note that if we did not know that $f\ '(a)$ existed, then the last line would not be valid, since $0$ times a nonexistent quantity is not defined. (In fact already the application of the Product Law for limits would not be valid.)