Question

Evaluate the following limits
$\underset{x\to 3}{lim}\frac{x^2-9}{x^2\left(x^2-6x+9\right)}$

Answer 1

$\underset{x\to 3}{lim}\frac{(x^2-9)}{x^2(x^2-6x+9)}$

$\Rightarrow \underset{x\to 3}{lim}\frac{(x^2-9)}{x^4-6x^3+9x^2}$

this is of the form $\frac{0}{0}$

$\therefore$ we can differentiate by using L'hospital rule

$\Rightarrow \underset{x\to 3}{lim}\left[\frac{\frac{d}{dx}(x^2-9)}{\frac{d}{dx}(x^4-6x^3+9x^2)}\right]$

$\Rightarrow \underset{x\to 3}{lim}\frac{2x}{4x^3-18x^2+18x}$

$\Rightarrow \frac{2\left(3\right)}{4\left(2^3\right)-18\left(2^2\right)+18\left(2\right)}$

$\Rightarrow \frac{6}{32-18(4-2)}$

$\Rightarrow \frac{6}{32-36}\Rightarrow \frac{-6}{4}\Rightarrow \frac{-3}{2}$

$\therefore \underset{x\to 3}{lim}\frac{(x^2-9)}{x^2(x^2-6x+9)}=\frac{-3}{2}$