To understand this, you need to think of the intuition behind the $\epsilon$-$\delta$ definition. We want $\lim_{x\to a}f(x)=L$ if we can make $f(x)$ as close to $L$ as we like by making $x$ sufficiently close to $a$. Worded differently, we might say that:
$\lim_{x\to a}f(x)=L$ if given any neighborhood $U$ of $L$, there is a neighborhood $V$ of $a$ such that elements of $V$ are mapped by $f$ to elements of $U$ (except possibly $a$ itself).
In this context, a "neighborhood" of a point $p$ should be understood to mean "points sufficiently close to $p$". Let's make that precise by defining what we mean by "close". For $\epsilon>0$ (assumed, but not required, to be very small) define
$$B(x,\epsilon):=\{y\,:\,|x-y|<\epsilon\},$$
the ball of radius $\epsilon$ about $x$. For our purposes, we say $U$ is a neighborhood of $x$ if $U=B(x,\epsilon)$ for some $\epsilon>0$. (The usual definition only requires that $U$ contains such a ball.) Assuming $\epsilon>0$ is very small, this agrees with our intuition of what closeness should mean. Now if we go back to our neighborhood "definition" of a limit, you should be able to think about it for a bit and convince yourself that it is equivalent to the usual definition.
How does this relate to the problem with infinity? Given that infinity is not a real number (and things like distance from infinity do not make sense), we must revise what it means to be "close" to infinity. So for $M>0$ (assumed this time to be very large) define
$$B(+\infty,M):=\{y\,:\,y>M\},\quad B(-\infty,M):=\{y\,:\,y<-M\},$$
the neighborhoods of $\pm\infty$. Hopefully you can see why these make sense as definitions; a number should be close to infinity if it is very large (with the correct sign), so a neighborhood of infinity should contain all sufficiently large numbers.
Now we extend our neighborhood definition of limits to include the case where $a$ or $L$ can be $\pm\infty$. It is a similar exercise to before to verify now that the definition is still equivalent to the old one, only now we have in some sense unified somewhat.
Let $|x| \lt 1/5$, then
1)$-1/5< x < 1/5$, or
$-3/5 +1< 3x +1<3/5+1$;
2) $|2x+3| \le 2|x| +3\lt 17/5$.
Let $\epsilon>0.$
Choose $\delta = \min(1/5, (2/17)\epsilon )$, then
$|x|\lt \delta$ implies
$\dfrac{|2x+3|}{|3x+1|}|x| \lt\dfrac{17/5}{2/5}|x| =$
$(17/2)\delta \lt \epsilon$.
Best Answer
Consider $$ f(x) = \begin{cases}x^2, & x \ne 0\\ 1, & x=0 \end{cases} $$
Do you want $\displaystyle \lim_{x\to 0} f(x)$ to exist? If you want it to exist and be zero, you need to rule out $x=c$ in the definition.