Question

Find the following limit:
$\underset{x\to 16}{lim}\frac{\sqrt[4]{4x}-2}{\sqrt{x}-4}.$

Answer 1

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

$=\underset{x\to 16}{lim}\frac{x^{\frac{1}{4}}-2}{x^{\frac{1}{2}}-2^2}$

$=\underset{x\to 16}{lim}\frac{x^{\frac{1}{4}}-2}{{\left(x^{\frac{1}{4}}\right)}^2-2^2}$

$\Rightarrow \underset{x\to 16}{lim}\frac{(x^{\frac{1}{4}}-2)}{(x^{\frac{1}{4}}+2)(x^{\frac{1}{4}}-2)}$

$\Rightarrow \underset{x\to 16}{lim}\frac{1}{x^{\frac{1}{4}}+2}$

$\Rightarrow \underset{x\to 16}{lim}\frac{1}{(16{)}^{\frac{1}{4}}+2}$

$\Rightarrow \frac{1}{2+2}=\frac{1}{4}$.