# Find the limits: lim_{xrightarrow2}frac{sqrt{x^2+12}-4}{x-2}

Limits and continuity
Find the limits:
$$\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}$$

2020-11-11

We have to find limits:
$$\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}$$
Since we have an indeterminate form of type $$\frac{0}{0}$$, we can apply the L'hospital rule:
$$\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}=\lim_{x\rightarrow2}\frac{\frac{d}{dx}(\sqrt{x^2+12}-4)}{\frac{d}{dx}(x-2)}$$
$$\lim_{x\rightarrow2}\frac{\frac{2x}{2\sqrt{x^2+12}}}{1}$$
$$\lim_{x\rightarrow2}\frac{x}{\sqrt{x^2+12}}=\frac{2}{\sqrt{2^2+12}}=\frac{2}{\sqrt{4+12}}$$
$$=\frac{2}{\sqrt{16}}=\frac{2}{4}=\frac{1}{2}$$
$$\Rightarrow\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}=\frac{1}{2}$$