Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist. I turned that into one fraction, then I factored the polynomial on the numerator as $-(x-1)(x+2)$ and the one on the denominator as $(x+1)(-x^2-x-1)$. What did I do wrong?

$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)=\lim_{x\to 1} \frac{x^2 + x + 4}{x^3 - 1}=\lim_{x\to 1}\frac{p(x)}{q(x)}$, where $p(x) =x^2+x+4, q(x) =x^3-1 $ 
Note that, $p(x) \to 6, q(x) \to 0$ as $x\to 1$
Now let's claim that the  limit $\lim_{x\to 1}\frac{p(x)}{q(x)}$ doesn't exist on set of real nos. $\mathbb R$. 
On the contrary, suppose that the above limit (i.e. $\lim_{x\to 1}\frac{p(x)}{q(x)}$) exists. Hence let$\lim_{x\to 1}\frac{p(x)}{q(x)}=L$, then we have:
$p(x) =\frac{p(x)} {q(x)} q(x) \implies \lim_{x\to 1} p(x) = L\times 0=0$, which is a contradiction. Hence the limit doesn't exist on set of real nos. and hence our claim is correct. 
Also, even if we consider extended set of real nos. i.e. $\mathbb R\cup \{-\infty, \infty\} $, then also left hand limit(LHL) $\lim_{x\to 1^{-}}\frac{p(x)}{q(x)}= - \infty$ and right hand limit(RHL) $\lim_{x\to 1^{+} }\frac{p(x)}{q(x)}=\infty$ and hence LHL and RHL are not equal and therefore, the limit doesn't exist on extended set of real nos. also.
