<s>[INST] How to show that $\lim_{x\to 1}(x^3-1)/(x-1)=3$? How to show that
$$
\lim_{x\to 1}\frac{x^3-1}{x-1}=3?
$$
I tried to solve but couldn't. Please help me.
 [/INST] Let $x = y+1$,
so we want to see what happens as $y \to 0$.
Note:
I almost always want to have
a variable that controls a limit
go to zero
instead of some other value
(like $1$ in this case).
\begin{align}
\frac{x^3-1}{x-1}
&=\frac{(y+1)^3-1}{(y+1)-1}\\
&=\frac{(y^3+3y^2+3y+1)-1}{y}\\
&=\frac{y^3+3y^2+3y}{y}\\
&=y^2+3y+3
\qquad\text{whenever }y \ne 0\\
&\to 3 \qquad\text{as } y \to 0\\
\end{align}
Note that this shows that
$\frac{x^3-1}{x-1}
=(x-1)^2+3(x-1)+3
$.
</s>