Proving the limit of a function of a sequence is equal to the function of the limit of that sequence

Just write down the definitions:

• $x_n$ converges to $c$ if and only if $\forall \delta > 0$ there exists $N = N(\delta)$ such that $\forall n \ge N$ we have $|x_n - c| < \delta$
• $f$ is continuous if and only if $\forall \eta > 0$ there exists $\gamma = \gamma(\eta)$ such that if $|x - y| < \gamma$ then $|f(x) - f(y)| < \eta$.

Now we want to prove the following claim

• $\forall \epsilon > 0$ there exists $M = M(\epsilon)$ such that if $n \ge M$ then we have $|f(x_n) - f(c)| < \epsilon$.

Hint: if $x_n \to c$ then you can make $|x_n - c|$ small enough to use the continuity of $f$ (say, for example, smaller than $\gamma(\epsilon)$).

Here is my attempt to give a complete proof, trying to stay consistent with the notation introduced by the previous answer.

On the one hand, continuity of f at c implies that $\forall \ \epsilon>0, \ \exists \ \gamma = \gamma(\epsilon)>0 \ s.t. \ |f(x_n) - f(c)| < \ \epsilon \ \forall \ x_n \ s.t. \ |x_n - c| < \ \gamma $.

On the other hand, convergence of $x_n$ to $c$ implies that $\exists$ $N=N(\gamma)$ s.t. $|x_n - c|<\gamma$ $\forall$ $n>N$.

Hence, $\exists$ $N=N(\gamma(\epsilon))$ s.t. $|x_n-c|<\gamma$ and, therefore (by continuity of f), $\ |f(x_n) - f(c)| < \ \epsilon$ for $n>N$.