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Calculate the following limit

Solution

$\lim_{x \rightarrow 4}\frac{2x^{3}-128}{\sqrt{x}-2}=\lim_{x \rightarrow 4}\frac{2\left(x^{3}-64\right)}{\sqrt{x}-2}$
$=\lim_{x \rightarrow 4}\frac{2\left(x^{3}-4^{3}\right)}{\sqrt{x}-2}$
$=\lim_{x \rightarrow 4}\frac{2\left(x-4\right)\left(x^{2}+4x+16\right)}{\sqrt{x}-2}$
$=\lim_{x \rightarrow 4}\frac{2\left(x-4\right)\left(x^{2}+4x+16\right)}{\sqrt{x}-2}\left(\frac{\sqrt{x}+2}{\sqrt{x}+2}\right)$
$=\lim_{x \rightarrow 4}\frac{2\left(x-4\right)\left(x^{2}+4x+16\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}$
$=\lim_{x \rightarrow 4}\frac{2\left(x-4\right)\left(x^{2}+4x+16\right)\left(\sqrt{x}+2\right)}{x-4}$
$=\lim_{x \rightarrow 4}2\left(x^{2}+4x+16\right)\left(\sqrt{x}+2\right)$
$=2\left(4^{2}+4\cdot4+16\right)\left(\sqrt{4}+2\right)$
$=2\left(16+16+16\right)\left(2+2\right)$
$=2\left(48\right)\left(4\right)$
$=384$
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