If a sequence $s_n$ converges, does $f(s_n)$ converge? Let $f:X \to Y$ be a function. Let $s_n$ be a convergent sequence in $X$. Then does the sequence $f(s_n)$ converge as well? Does $f$ have to be continuous?
My thoughts: 
I believe $f$ does converge. If $s_n \to s$ then applying $f$ to both sides yields $f(s_n)\to f(s).$ Hence $f(s_n)$ converges. Is it that simple?
Now assume $f$ is not continuous. Then $\displaystyle\lim_{x\to t} f(x) \ne f(t).$ But if I take a limit point $l \in X$, then $\exists$ $l_n\in X-\{l\}$ such that $l_n \to l.$ If I apply $f$ to both sides, I have $f(l) \to f(l).$ So $f$ converges $\implies$ Contradiction! $\implies f$ is continuous. 
Please let me know what you think or if you can answer my questions. Thanks a lot!
 A: I believe that "applying $f$ to both sides" is dependent on continuity. Consider the function $f(x) = \begin{cases} 1, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Clearly the sequence $1, 1/2, 1/4, \ldots$ converges to $0$. But applying $f$ to it gives $1, 1, 1, \ldots$, which does not converge to $f(0) = 0$.
If $s_n$ converges to $s$ and $f$ is continuous at $x$, then $f(s_n)$ converges to $f(s)$.
Let $\epsilon > 0$. Because $f$ is continuous at $s$, there is a $\delta > 0$ such that if $|x - s| < \delta$, then $|f(x) - f(s)| < \epsilon$. Because $s_n$ converges to $s$, there is an $N \in \mathbb{N}$ such that $|s_i - s| < \delta$ for all $i > N$. Thus, for all $i > N$, $|f(s_i) - f(s)| < \epsilon$, and so the sequence $f(s_n)$ converges to $f(s)$.
A: Let $X=Y=\mathbb R$.
Regarding the second question, consider $f\colon X\to Y$ defined by $f(x)=0$ if $x\neq 0$ and $f(0)=1$.
Clearly $f$ isn't continuous. Does $\left(f\left( \dfrac 1 n\right)\right)_{n\in \mathbb N}$ converge?
About the first one, consider the function $f\colon X\to Y$ (which isn't continuous), defined by $$f(x)=\begin{cases} 0, &\text{ if }x\in \mathbb Q\\
1, &\text{ if }x\in \mathbb R\setminus \mathbb Q\end{cases}$$
Does $\left(f\left( \dfrac 1 {\sqrt n}\right)\right)_{n\in \mathbb N}$ converge?
