Question

Solve : $\underset{x\to 1}{lim}\frac{x^3-3x+1}{x-1}$

Answer 1

We have, $\underset{x\to 1}{lim}\frac{x^3-3x+1}{x-1}$

Let $L=\underset{x\to 1}{lim}\frac{x^3-3x+1}{x-1}$

This is $\frac{0}{0}$ form.

So, apply L-Hospital rule,

$=\underset{x\to 1}{lim}\frac{3x^2-3+0}{1-0}$

$=\underset{x\to 1}{lim}\frac{3x^2-3}{1}$

$=\underset{x\to 1}{lim}3x^2-3$

$=3(1{)}^2-3$

$L=3-3=0$

Hence, this is the answer.