Question

Evaluate the limit:
$\underset{x\to 4}{lim}\frac{x^2-16}{\sqrt{x}-2}$

Answer 1

Given: $\underset{x\to 4}{lim}\frac{x^2-16}{\sqrt{x}-2}=becomes\frac{0}{0}form$

Using Factorization method

$\underset{x\to 4}{lim}\frac{x^2-16}{\sqrt{x}-2}$

$\underset{x\to 4}{lim}\frac{(x-4)(x+4)}{\sqrt{x}-2}$

$[\because (a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

$\underset{x\to 4}{lim}\frac{({\left(\sqrt{x}\right)}^2-2^2)(x+4)}{\sqrt{x}-2}$

$\underset{x\to 4}{lim}\frac{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)}{\sqrt{x}-2}$

$\underset{x\to 4}{lim}(\sqrt{x}+2)(x+4)$

$=(\sqrt{4}+2)(4+4)$

$=4\left(8\right)=32$

$\therefore \underset{x\to 4}{lim}\frac{x^2-16}{\sqrt{x}-2}=32$