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Q. $\displaystyle\lim _{x \rightarrow 4}\left[\frac{x^{3 / 2}-8}{x-4}\right]$ equals

KEAMKEAM 2019

Solution:

$y=\displaystyle\lim _{x \rightarrow 4}\left[\frac{x^{3 / 2}-8}{x-4}\right]=\displaystyle\lim _{x \rightarrow 4}\left[\frac{\left(x^{1 / 2}\right)^{3}-(2)^{3}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right]$
$y=\displaystyle\lim _{x \rightarrow 4}\left[\frac{\left(x^{1 / 2}-2\right)(x+4+2 \sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)}\right]$
$y=\displaystyle\lim _{x \rightarrow 4} \frac{(x+4+2 \sqrt{x})}{(\sqrt{x}+2)}=\frac{12}{4}=3$