Evaluate the following limits: _ x 0 [ 6+x+x^2-6 x ] - Vidyadip

Question

Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt{6}}{x}\right]$

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Answer

$ \lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt{6}}{x}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt{6}}{x} \times \frac{\sqrt{6+x+x^2}+\sqrt{6}}{\sqrt{6+x+x^2}+\sqrt{6}}\right] $
...[By rationalization]
$ =\lim _{x \rightarrow 0} \frac{\left(6+x+x^2\right)-6}{x\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\lim _{x \rightarrow 0} \frac{x+x^2}{x\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\lim _{x \rightarrow 0} \frac{x(1+x)}{x\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\lim _{x \rightarrow 0} \frac{1+x}{\sqrt{6+x+x^2}+\sqrt{6}} \quad \ldots[\because x \rightarrow 0, x \neq 0]$
$=\frac{\lim _{x \rightarrow 0}(1+x)}{\lim _{x \rightarrow 0}\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\frac{(1+0)}{\sqrt{6}+\sqrt{6}}$
$=\frac{1}{2 \sqrt{6}} $

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