Define
$f:(-1,1)\rightarrow\mathbb{R}$ by $f(x)=\frac{x+1}{x^2-1}$. Does $f$ have a limit at 1?
Well we know that $f(x)=\frac{x+1}{x^2-1}=\frac{1}{x-1}$. Thus clearly there is no limit.
By definition,
$\lim_{x\rightarrow c}f(x)$ means $ \forall N>0:\exists \delta :0<|x-c|<\delta \implies f(x)>N$.
So, my attempt/combined with answer given:
Let $\epsilon>0 \text{ and } L \in \mathbb{R}$. Then, $|f(x)-L|<\epsilon$, for $0<x-1<\frac{1}{L+\epsilon}$ (I don't understand this part). Thus, $L+\epsilon>\frac{1}{L+\epsilon}=f(x)$. Hence, $|f(x)-L|>\epsilon$. Therefore, it is impossible to find $\delta>0$ that fulfills the definition. Thus, $f$ does not have a limit at $x=1$.
Can someone perhaps show me a better way of doing this?
Using the definition as directly as possible will help your proof.
Let me work with the function $g(x) = \frac1{|x-1|}$ for now; we'll prove that $\lim_{x\to1}g(x) = \infty$.
So, take any $N > 0$. Our goal is to find a $\delta$ such that if $0 < |x-1| < \delta$, then $g(x) > N$. Note that by taking reciprocals of the inequality $|x-1| < \delta$, we get $$\frac1{|x-1|}>\frac1\delta.$$ The left-hand side is $g(x)$, so we're almost done. We just need to find a $\delta$ that works; in this case, pick any $\delta < \frac1N$, for then $$g(x)=\frac1{|x-1|}>\frac1\delta>N.$$
This gives you almost everything you need to prove the limit for your function $f(x)$.