Pick out the correct choices -TIFR 2015 Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function and $A \subset \mathbb R$ be defined by $A=\{y \in \mathbb R:y=\lim _{n\rightarrow \infty}f(x_n),$for some sequence  $x_n\rightarrow \infty\}$ 
Then the set $A$ is necessarily 
A.a connected set
B.compact set
C. a singleton set 
D.None of above
Now since $f$ is a continuous function and $x_n$ diverges so $f(x_n)$ will also diverge.Hence the set can't be bounded and hence not compact 
Also it may not be singleton as $f(x_n)$ may be either $\infty$ or -$\infty$
Not sure with A.Can someone please check my solution and suggest required edits
 A: Continuous functions take convergent sequences to convergent sequences, but you can't say the same for divergent sequences. For instance, consider a constant function. For a better example, which will also give a hint for how to solve your problem, take $f(x) = \sin(x)$. For a still better example, think of how you could modify this $f$ to change the set $A$...
A: The answer is connected,as far as I think.Because take the function $f(x)=e^xsinx$ where $f:\mathbb R \to \mathbb R$ ,for this function the set $A$ is neither compact nor singleton.Again consider $f$ which is continuous.Let $y_1<y_2\in A$,then let $y$ be in between them.Now there exists $(a_n),(b_n)\to \infty$ such that $f(a_n)=y_1$ and $f(b_n)=y_2$.Now since $y_1<y_2$ then there exists $N$ such that $f(a_n)<y<f(b_n)$ for all $n>N$.Now there exists a sequence $(c_n)$ such that $a_n<c_n<b_n \forall n>N$,and $f(c_n)=y\forall n>N$.Then $\operatorname{lim}f(c_n)=y$ and by sandwich theorem $c_n \to \infty$ as $n\to \infty$.So $y\in A$.
