Question

${lim}_{x\to 1}\left(\frac{3x-3}{3x^2+3x-6}\right)$ equal to

Answer 1

We have,

$\underset{x\to 1}{lim}\left(\frac{3x-3}{3x^2+3x-6}\right)$

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

So, apply L-Hospital rule,

$\underset{x\to 1}{lim}\left(\frac{3-0}{6x+3-0}\right)$

$\underset{x\to 1}{lim}\left(\frac{3}{6x+3}\right)$

$=\frac{3}{6\times 1+3}$

$=\frac{1}{3}$

Hence, this is the answer.