Which of the following sets of functions from $\mathbb{R}$ to $\mathbb{R}$ is a vector subspace of $\mathbb{R}$? 
Which of the following sets of functions from $\mathbb{R}$ to $\mathbb{R}$ is a vector subspace of $\mathbb{R}$?
$S_1=\{f|\displaystyle\lim_{x\to 3}f(x)=0\}$, $S_2=\{h|\displaystyle\lim_{x\to 3}h(x)=1\}$, $S_3=\{g|\displaystyle\lim_{x\to 3}g(x)~ \text{exists}\}$ 
  
  
*
  
*Only $S_1$.
  
*Only $S_2.$
  
*$S_1$ and $S_3$ but not $S_2$
  
*All of the three are vector spaces.
  

$S_1$ is definitely a vector subspace but $S_2$ is not as it doesn't contain $0$. But I am confused about the $S_3$. As there is no specific limit mentioned, so it means all sorts of limits are possible, including $0$. So I guess $S_3$ is also a vector subspace. That means option 3 is correct.  Is my solution correct? Any help would be great. Thank you.

Your solution is correct.
For $S_3$ : Take $g,h \in S_3$.
Then $\lim_{x \to 3} g(x)$ and $\lim_{x \to 3} h(x)$ both exist.
Now check


*

*$\lim_{x \to 3} (g(x)+h(x))=\lim_{x \to 3} g(x)+\lim_{x \to 3} h(x)=\text{exists}.$ This implies that $g+h \in S_3$.

*let $a \in \Bbb R$ be arbitrary. Then $\lim_{x \to 3} ag(x)=a\lim_{x \to 3} g(x)=\text{exists}$. This implies $ag \in S_3$.


From 1 and 2, $S_3$ is also a vector space.
Note : "Limit exists" means that the limit is finite.
