Question

Evaluate the following limit, if it exists :
$\sf \lim_{x \to 4} \rm{ \dfrac{ {2x}^{3} - 128 }{ \sqrt{x} - 2}}$

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Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given limit.

$\displaystyle \rm =\lim_{x \to4} \bigg( \dfrac{2 {x}^{3} - 128}{ \sqrt{x} - 2 } \bigg)$

If we put x = 4, we will get indeterminate form. So, we will first transform the numerator.

$\displaystyle \rm =\lim_{x \to4} \bigg( \dfrac{2 ({x}^{3} - 64)}{ \sqrt{x} - 2 } \bigg)$

$\displaystyle \rm =\lim_{x \to4} \dfrac{2 ({x}^{3} - {4}^{3} )}{ \sqrt{x} - 2 }$

$\displaystyle \rm =\lim_{x \to4} \dfrac{2 (x - 4)( {x}^{2} + 4x + 16)}{ \sqrt{x} - 2 }$

$\displaystyle \rm =\lim_{x \to4} \dfrac{2 ( {( \sqrt{x}) }^{2} - {2}^{2} )( {x}^{2} + 4x + 16)}{ \sqrt{x} - 2 }$

$\displaystyle \rm =\lim_{x \to4} \dfrac{2 ( \sqrt{x} + 2)( \sqrt{x} - 2 )( {x}^{2} + 4x + 16)}{ \sqrt{x} - 2 }$

$\displaystyle \rm =\lim_{x \to4} 2 ( \sqrt{x} + 2)( {x}^{2} + 4x + 16 )$

$\displaystyle \rm =2 ( \sqrt{4} + 2)( {4}^{2} + 4 \times 4+ 16 )$

$\displaystyle \rm =2 \times 4 \times 48$

$\displaystyle \rm =384$

Therefore:

$\displaystyle \rm \longrightarrow \lim_{x \to4} \bigg( \dfrac{2 {x}^{3} - 128}{ \sqrt{x} - 2 } \bigg) = 384$

$\textsf{\large{\underline{Learn More}:}}$

$\displaystyle\rm 1.\:\: \lim_{x\to0}\sin(x)=0$

$\displaystyle\rm 2.\:\: \lim_{x\to0}\cos(x)=1$

$\displaystyle\rm 3.\:\: \lim_{x\to0}\dfrac{\sin(x)}{x}=1$

$\displaystyle\rm 4.\:\: \lim_{x\to0}\dfrac{\tan(x)}{x}=1$

$\displaystyle\rm 5.\:\: \lim_{x\to0}\dfrac{1-\cos(x)}{x}=0$

$\displaystyle\rm 6.\:\: \lim_{x\to0}\dfrac{\sin^{-1}(x)}{x}=1$

$\displaystyle\rm 7.\:\: \lim_{x\to0}\dfrac{\tan^{-1}(x)}{x}=1$

$\displaystyle\rm 8.\:\: \lim_{x\to0}\dfrac{\log(1+x)}{x}=1$

$\displaystyle\rm 9.\:\: \lim_{x\to0}\dfrac{e^{x}-1}{x}=1$