Evaluate the following limits: _ x 3 [ 2 x+3-4 x-3 x^2-9 ] - Vidyadip

Question

Evaluate the following limits: $\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^2-9}\right]$

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Answer

$\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^2-9}\right]$
$= \lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^2-9} \times \frac{\sqrt{2 x+3}+\sqrt{4 x-3}}{\sqrt{2 x+3}+\sqrt{4 x-3}}\right]......[$By rationalization$]$
$=\lim _{x \rightarrow 3}\left[\frac{(2 x+3)-(4 x-3)}{\left(x^2-9\right)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{-2 x+6}{\left(x^2-9\right)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{-2(x-3)}{(x+3)(x-3)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{-2}{(x+3)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$\text {}[x x \rightarrow 3 \neq 0 ; \neq 0$
$=\frac{-2}{(3+3)(\sqrt{2(3)+3}+\sqrt{4(3)-3})}$
$=\frac{-2}{6(3+3)}$
$=\frac{-1}{18}$

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