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

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