Evaluate $\lim_{x \to 1} \cfrac{x^3 - 1}{x^2 - 2x + 1}$ without l'Hospital

Question

I'm pretty stuck beyond identifying $$\lim_{x \to 1} \cfrac{x^3 - 1}{x^2 - 2x + 1} = \lim_{x \to 1} \cfrac{x^3 -1}{(x-1)^2}$$

Any hints? (Preferably not full answer)

Answer 1

Hint: $x^3-1=(x-1)(x^2+x+1)$.

Answer 2

$$ (x^3-1)=(x-1)(x^2+x+1). $$ $$ x^2-2x+1=(x-1)^2. $$ Then, $$ \lim_{x\rightarrow 1}\frac{x^2+x+1}{x-1}=? $$

Answer 3

It is $(x+2)+\frac{3}{x-1}$ see that the second term diverges to infinity.

Answer 4

Hint:

\begin{align} \lim_{x \to 1} \cfrac{x^3 - 1}{x^2 - 2x + 1} &= \lim_{x \to 1} \left(x+2+\cfrac{3}{x-1}\right) =\dots \end{align}