Question

$\underset{x\to 3}{lim}\frac{\left[\right(x^2-9\left)\right](x^3+27)}{ln(x-2)}$ is eqaul to

Answer 1

$x^3+27=(x+3)(x^2+9-3x)$
$x^2-9=(x-3)(x+3)$
so, $\underset{x\to 3}{lim}\frac{(x^3+27)\left(ln\right(x-2)}{(x^2-9)}$
$=\underset{x\to 3}{lim}\frac{(x^2+9-3x)ln(x-2)}{x-3}$
$=\underset{x\to 3}{lim}(x^2+9-3x)\cdot lim\frac{ln(x-2)}{(x-3)}$
$=(9+9-9)\cdot \underset{x\to 3}{lim}\frac{ln(x-2)}{-3}$
Putting $x=3+h$
$9\underset{x\to o}{lim}\frac{ln(1+h)}{h}$
this limit $=1$
$\Rightarrow 9\times 1=9$