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?
 A: Using the identity
$A-B=\dfrac{1}{\dfrac{1}{A}}-\dfrac{1}{\dfrac{1}{B}}$ we get $$\dfrac{1}{x-1}-\dfrac{3}{1-x^3}=\dfrac{2-x^3-x}{-x^4+x^3+x-1}$$ so you go from the form $\infty-\infty$ to the form $\dfrac00$ and you can apply L'Hôspital so you have $$\lim_{x \to 1} \frac{1}{x-1}-\frac{3}{1-x^3}=\lim\dfrac{-3x^2-1}{-4x^3+3x^2+1}$$ and you can verify that the limits to the right and to the left are not equal, the first is $-\infty$ and the second is $\infty$.
Thus the limit does not exist.
A: $$\frac1{x-1}-\frac3{1-x^3}=\frac1{x-1}+\frac3{x^3-1}\to\infty$$
as the terms do not cancel each other (they have the same sign).

Assuming a typo,
$$\frac1{x-1}+\frac3{1-x^3}=\frac1{x-1}-\frac3{x^3-1}=\frac{x^2+x-2}{x^3-1}=\frac{x+2}{x^2+x+1}\to1.$$
A: Simplifying the function as (1/(x-1))*(1+3/(x²+x+1))= f(x)g(x),where f(x)=(1/(x-1)) and g(x)=(1+3/(x²+x+1)).
Clearly, limit value of g(x) at x tends to 1 becomes finite.
But about f(x), we can't say that,,, clearly f(x) tend to ∞ as x tend to 1 from the right side i.e x tend to 1+, on the other hand, f(x) tend to -∞ as x tends to 1 from the left side i.e x tends to 1-, limit of f doesn't exist (even in infinitely) as x tends to 1.
Also you can think through the graph of 1/(x-1), it will give much more understanding to you.
A: $\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)=\lim_{x\to 1} \frac{x^2 + x + 4}{x^3 - 1}=\lim_{x\to 1}\frac{p(x)}{q(x)}$, 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.
A: I believe you messed up by looking at $$\lim_{x \to 1}\left(\frac{1}{x-1} \color{red}{+} \frac{3}{1-x^3}\right)$$ instead of $$\lim_{x \to 1} \left( \frac{1}{x-1}\color{red}{-}\frac{3}{1-x^3} \right)$$
The value of the first one is $1$, but the second one does not exist.
We can factor $1-x^3$ as $(x-1)(-x^{2}-x-1)$ so that the limit becomes $$\lim_{x \to 1}\frac{(-x^2-x-1)-3}{(x-1)(-x^2-x-1)}$$
Then it is clear that the limit does not exist since the numerator when plugging in $x = 1$ is $-6$ and the denominator is $0$.
