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?
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.
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 $0$.