Question

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Answer 1

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

$\sf \: Evaluate \: \displaystyle\lim_{x \to 0}\sf \dfrac{ \sqrt[5]{x + 32} - 2}{x}$

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

Given function is

$\rm :\longmapsto\: \displaystyle\lim_{x \to 0}\sf \dfrac{ \sqrt[5]{x + 32} - 2}{x}$

If we substitute x = 0, directly we get

$\rm \:  =  \:  \: \ \dfrac{ \sqrt[5]{0 + 32} - 2}{0}$

$\rm \:  =  \:  \: \ \dfrac{ \sqrt[5]{ 32} - 2}{0}$

$\rm \:  =  \:  \: \ \dfrac{ 2 - 2}{0}$

$\rm \:  =  \:  \: \ \dfrac{0}{0} \: which \: is \: meaningless$

So, To evaluate this limit,

$\rm :\longmapsto\: \displaystyle\lim_{x \to 0}\sf \dfrac{ \sqrt[5]{x + 32} - 2}{x}$

we use method of Substitution.

$\red{\rm :\longmapsto\:Put \: \sqrt[5]{x + 32} = y}$

$\red{\rm :\longmapsto\:x + 32 = {y}^{5} }$

$\red{\rm :\longmapsto\:x = {y}^{5} - 32 }$

$\red{\rm :\longmapsto\:When \: x \to \: 0 \: \: so \: \: y \: \to \: 2}$

So, the given

$\rm :\longmapsto\: \displaystyle\lim_{x \to 0}\sf \dfrac{ \sqrt[5]{x + 32} - 2}{x}$

reduces to

$\rm \:  =  \:  \: \displaystyle\lim_{y \to 2}\sf \: \: \frac{y - 2}{ {y}^{5} - 32}$

can be rewritten as

$\rm \:  =  \:  \: \displaystyle\lim_{y \to 2}\sf \: \: \frac{y - 2}{ {y}^{5} - {2}^{5} }$

can be again rewritten as

$\rm \:  =  \:  \: \displaystyle\lim_{y \to 2}\sf \: \: \dfrac{1}{\dfrac{ {y}^{5} - {2}^{5} }{y - 2} }$

We know that,

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

So, using this, we get

$\rm \:  =  \:  \: \dfrac{1}{5 {(2)}^{5 - 1} }$

$\rm \:  =  \:  \: \dfrac{1}{5 {(2)}^{4} }$

$\rm \:  =  \:  \: \dfrac{1}{5 \times 16}$

$\rm \:  =  \:  \: \dfrac{1}{80}$

Hence,

$\purple{\boxed{ \bf{ \: \: \displaystyle\lim_{x \to 0}\bf \dfrac{ \sqrt[5]{x + 32} - 2}{x} = \frac{1}{80} }}}$

Additional Information :-

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

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

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

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

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