<s>[INST] Getting the correct answer for a limit with a cube in the denominator Given:
$$
\lim_{x \to 1} f(x) = \frac3{1-x^3} - \frac1{1-x}
$$
I first used the difference of cubes to get
$$
\lim_{x \to 1} f(x) = \frac3{(1-x)(1+x+x^2)} - \frac1{1-x}
$$
Then multiplied each term by $(1-x)$, cancelling $(1-x)$ in the first term, and making the second term $1$
$$
\lim_{x \to 1} f(x) = \frac3{(1+x+x^2)} - 1
$$
After substitution 
$$
\lim_{x \to 1} f(x) = \frac33 - 1
$$
Which equates to $0$.  But I know the correct answer is $1$.  Where did I slip up?
 [/INST] You can't multiply each term by $(1-x)$- rather you should factor it out to get:
$\frac{1}{1-x}\frac{3}{(1+x+x^2)-1)}=\frac{1}{1-x}\frac{(2-x-x^2)}{(1+x+x^2)}=\frac{1}{1-x}\frac{(1-x+1-x^2)}{(1+x+x^2)}=\frac{1}{1-x}\frac{(1-x)(2+x)}{(1+x+x^2)}$. 
Now, cancel, and take the limit- the answer shall be $1$
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