Question

Can anyone solve this? send me correct solution ​

Answer 1

Given Question :-

Evaluate the following limit :-

$\displaystyle \lim_{x \to \: 16} \frac{ {\bigg(x \bigg) }^{\dfrac{1}{4} } - 2 }{{\bigg(x\bigg) }^{\dfrac{1}{2} } - 4}$

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

Given expansion is

$\rm :\longmapsto\:\displaystyle \lim_{x \to \: 16} \frac{ {\bigg(x \bigg) }^{\dfrac{1}{4} } - 2 }{{\bigg(x\bigg) }^{\dfrac{1}{2} } - 4}$

$\rm \:  =  \: \dfrac{ {\bigg(16 \bigg) }^{\dfrac{1}{4} } - 2 }{{\bigg(16\bigg) }^{\dfrac{1}{2} } - 4}$

$\rm \:  =  \: \dfrac{2 - 2}{4 - 4}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, we use Method of Substitution to solve this limit

$\rm :\longmapsto\:\displaystyle \lim_{x \to \: 16} \frac{ {\bigg(x \bigg) }^{\dfrac{1}{4} } - 2 }{{\bigg(x\bigg) }^{\dfrac{1}{2} } - 4}$

So, Substitute

$\red{\rm :\longmapsto\:{\bigg(x\bigg) }^{\dfrac{1}{4} } = y \: \rm \implies\:x = {y}^{4}}$

$\red{\rm :\longmapsto\:as \: x \: \to \: 16, \: \: so \: y \: \to \: 2 \: }$

So, above expression can be rewritten as

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

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

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

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

$\rm \:  =  \: \dfrac{1}{2 + 2}$

$\rm \:  =  \: \dfrac{1}{4}$

Hence,

$\rm \implies\: \: \: \boxed{ \tt{ \: \displaystyle \lim_{x \to \: 16} \frac{ {\bigg(x \bigg) }^{\dfrac{1}{4} } - 2 }{{\bigg(x\bigg) }^{\dfrac{1}{2} } - 4} = \frac{1}{4} \: }}$

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Additional Information :-

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

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

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

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

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