Question

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

Answer 1

$$\begin{gathered} \underset{x\to 4}{\lim }\left[\frac{x^2-16}{\sqrt{x}-2}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to 4}{\lim }\left[\frac{x^2-4^2}{\sqrt{x}-2}\right] \\ =\underset{x\to 4}{\lim }\left[\frac{\left(x-4\right)\left(x+4\right)}{\left(\sqrt{x}-2\right)}\right] \\ =\underset{x\to 4}{\lim }\left[\frac{\left\{{\left(\sqrt{x}\right)}^2-2^2\right\}\left(x+4\right)}{\left(\sqrt{x}-2\right)}\right] \\ =\underset{x\to 4}{\lim }\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(x+4\right)}{\left(\sqrt{x}-2\right)}\right] \\ =\left(2+2\right)\left(4+4\right) \\ =32 \end{gathered}$$