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.
 A: If $x\neq 1$, then (this is a geometric sum)
$$
1+x+x^2=\frac{x^3-1}{x-1}.
$$
I guess you can proceed from here?
Edit
Although I think that it is good to know and recognize geometric series, here are two other ways.
1) L'Hospital.
$$
\lim_{x\to 1}\frac{x^3-1}{x-1}=\lim_{x\to 1}\frac{3x^2}{1}=3.
$$
2) Since the polynomial $f(x)=x^3-1$ satisfies $f(1)=0$, we know by the factor theorem that $(x-1)$ is a factor of $f(x)$. Long polynomial division gives
$$
\frac{x^3-1}{x-1}=x^2+x+1.
$$
Then, the limit is easily seen to be $3$.
3) The function $f(x)=x^3$ is differentiable everywhere, with derivative $f'(x)=3x^2$. By the definition of derivative
$$
3=f'(1)=\lim_{x\to 1}\frac{x^3-1}{x-1}.
$$
Choose any way, but I still like the one with geometric series best.
A: Hint $x^3-1=(x-1)(x^2+x+1)$ and $x$ is not exactly $1$.
A: The fact that $x^3-1$ becomes $0$ when $x=1$ tells you that $x-1$ is a factor of $x^3-1$.  So you have
$$
x^3-1 = (x-1)(\cdots\cdots).
$$
You can find the other factor by long division, then cancel $x-1$ from the numerator and the denominator.  After that it's easy.
Basic fact from algebra.  If you plug a number (e.g. $4$) into a polynomial and get $0$, then $x$ minus that number (e.g. $x-4$) is a factor of that polynomial.
A: 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
$.
A: You can use polynomial long division (https://en.wikipedia.org/wiki/Polynomial_long_division) on this problem
$$(x^3-1):(x-1)=x^2+x+1$$
Then taking the limit is reduced to setting $x=1$ into $x^2+x+1$. 
