Question

Please solve the question in attachment​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Evaluate the following limit

$\rm :\longmapsto\:\displaystyle\lim_{x \to 64} \frac{{\bigg[x\bigg]}^{\dfrac{1}{6}} - 2}{{\bigg[x\bigg]}^{\dfrac{1}{3}} - 4}$

$\red{\large\underline{\sf{Solution-}}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 64} \frac{{\bigg[x\bigg]}^{\dfrac{1}{6}} - 2}{{\bigg[x\bigg]}^{\dfrac{1}{3}} - 4}$

If we substitute directly x = 64, we get

$\rm \:  =  \: \dfrac{{\bigg[64\bigg]}^{\dfrac{1}{6}} - 2}{{\bigg[64\bigg]}^{\dfrac{1}{3}} - 4}$

$\rm \:  =  \: \dfrac{{\bigg[ {2}^{6} \bigg]}^{\dfrac{1}{6}} - 2}{{\bigg[ {4}^{3} \bigg]}^{\dfrac{1}{3}} - 4}$

$\rm \:  =  \: \dfrac{2 - 2}{4 - 4}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, to evaluate the

$\rm :\longmapsto\:\displaystyle\lim_{x \to 64} \frac{{\bigg[x\bigg]}^{\dfrac{1}{6}} - 2}{{\bigg[x\bigg]}^{\dfrac{1}{3}} - 4}$

we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = {y}^{6}, \: as \: x \to \: 64, \: \: so \: y \to \: 2 \: }$

So, above expression can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{y \to 2} \frac{{\bigg[ {y}^{6} \bigg]}^{\dfrac{1}{6}} - 2}{{\bigg[ {y}^{6} \bigg]}^{\dfrac{1}{3}} - 4}$

$\rm \:  =  \: \displaystyle\lim_{y \to 2} \frac{y - 2}{ {y}^{2} - 4}$

$\rm \:  =  \: \displaystyle\lim_{y \to 2} \frac{y - 2}{ {y}^{2} - {2}^{2} }$

$\rm \:  =  \: \displaystyle\lim_{y \to 2} \frac{\cancel{y - 2}}{\cancel{(y - 2)} \: \: (y + 2)}$

$\rm \:  =  \: \displaystyle\lim_{y \to 2} \frac{1}{y + 2}$

$\rm \:  =  \: \dfrac{1}{2 + 2}$

$\rm \:  =  \: \dfrac{1}{4}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 64} \frac{{\bigg[x\bigg]}^{\dfrac{1}{6}} - 2}{{\bigg[x\bigg]}^{\dfrac{1}{3}} - 4} = \frac{1}{4} \: }}$

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More to Know :-

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \: \frac{sinx}{x} \: = \: 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \: \frac{tanx}{x} \: = \: 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \: \frac{log(1 + x)}{x} \: = \: 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \: \frac{ {e}^{x} \: - \: 1 }{x} \: = \: 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \: \frac{ {a}^{x} \: - \: 1 }{x} \: = \: loga \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to a} \: \frac{ {x}^{n} - {a}^{n} }{x - a} \: = \: {na}^{n - 1} \: }}$

Answer 2

hope you went correct answer from my side