Q:

Convergent strictly increasing sequence $a_n$ $\Rightarrow$ sequence $f(a_n)$ is convergent.

Strictly increasing $f(x)$ is defined on R. Then for any convergent
  strictly increasing sequence $a_n$ sequence $f(a_n)$ is also
  convergent.

The answer is TRUE, but I believe I have a counterexample:
$a_n = 1-\frac{1}{n}; f(x) = x+n \Rightarrow f(a_n) = 1 - \frac{1}{n} + n$
Am I correct?

A:

$f$ is strictly increasing function. $a_n$ is a strictly increasing convergent sequence, then $a_n$ is bounded. Let $|a_n|\le M \quad \forall n\in \mathbb{N}$, then the sequence $f(a_n)$ is strictly increasing, and $|f(a_n)|\le |f(M)|$; i.e $f(a_n)$ is bounded and hence converges by monotone convergence theorem.

That means $\lim_{n\rightarrow \infty}{f(a_n)}$ exists.  Now a possible question is, if $a_n\rightarrow l$ whether the follows is true: $$\lim_{n\rightarrow \infty}{f(a_n)}=f(l) \tag{1}$$
No. Consider the function $$ f(x)= \left\{\begin{array}{cl} e^x, &x\lt 0 \\ e^x+1&x \ge 0 \end{array}\right.  $$ Clearly $f$ is strictly increasing, has jump discontinuity at $x=0$.
Let $a_n=-\frac{1}{n}$ now $a_n$ is strictly increasing, converges to $0$. But $f(a_n)=1, n\ge 1$ while $f(0)=2$.
That said, if $f$ is continuous $(1)$ holds.

