Show that $(x_n)_{n\in\mathbb{N}}$ converges to $x$ if and only if $(f(x_n))_{n\in\mathbb{N}}$ converges to $f(x)$, where f is strictly increasing. We are supposed to use a previously solved problem to help deduce this, but I can't figure out how they relate. The other problem goes as follows: "Suppose $g$ is a strictly increasing function and let $B$ be the range of $g$. Prove that the inverse of $g$ is continuous."
I managed to solve that part, but I don't know how to use it to solve the rest. Any help would be appreciated.
 A: After some more digging, and consulting some other sources, I think I have a sufficient answer. 
For the reverse direction, we assume $f$ is continuous at $x$, and that $x_n$ approaches $x.$
Let $\varepsilon>0.$
We need to find a $N \in \mathbb{N}$ such that $|f(x_n)-f(x)| <\varepsilon$ for every $n \geq N.$
Since $f$ is continuous at $x$, there exists a $\delta >0$ such that $|f(y)-(fx)| <\varepsilon$ whenever $|y-x|<\delta.$
Since $x_n \rightarrow x$, there exists $N \in \mathbb{N}$ such that $|x_n-x|< \delta$ for every $n \geq N$. We can use this $N$ in our definition above.
Therefore, $|f(x_n)-f(x)|< \varepsilon$ for every $n \geq N.$
For the forwards direction, we assume towards a contradiction that $f$ is not continuous at $x.$
Then, for some $\varepsilon >0$, and for each $n \geq N,$ there exists a $x_n$ such that $|x_n-x|<\frac{1}{n}$, but $|f(x_n)-f(x)|\geq \varepsilon.$ 
This is a contradiction, because we are claiming that $x_n \rightarrow x$ if and only if $f(x_n) \rightarrow f(x).$
And that's the proof. I could be wrong, but to me it seems to make sense.
