Question

#Quality Question

#Limits

See Attachment

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Answer 1

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

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \bigg(\dfrac{ {3x}^{2} + 2}{ {7x}^{2} + 2}\bigg)^{\dfrac{1}{ {x}^{2} } }$

On adding and Subtracting 1, we get

$\rm \: = \:\displaystyle\lim_{x \to 0}\rm \bigg(1 + \dfrac{{3x}^{2} + 2}{ {7x}^{2} + 2} - 1\bigg)^{\dfrac{1}{ {x}^{2} } }$

$\rm \: = \:\displaystyle\lim_{x \to 0}\rm \bigg(1 + \dfrac{{3x}^{2} + 2 - ( {7x}^{2} + 2 )}{ {7x}^{2} + 2} \bigg)^{\dfrac{1}{ {x}^{2} } }$

$\rm \: = \:\displaystyle\lim_{x \to 0}\rm \bigg(1 + \dfrac{{3x}^{2} + 2 - {7x}^{2} - 2 }{ {7x}^{2} + 2} \bigg)^{\dfrac{1}{ {x}^{2} } }$

$\rm \: = \:\displaystyle\lim_{x \to 0}\rm \bigg(1 + \dfrac{ - {4x}^{2}}{ {7x}^{2} + 2} \bigg)^{\dfrac{1}{ {x}^{2} } }$

$\rm \: = \:\displaystyle\lim_{x \to 0}\rm \bigg(1 + \dfrac{ - {4x}^{2}}{ {7x}^{2} + 2} \bigg)^{\dfrac{ - 4 {x}^{2} }{ {x}^{2} ( {7x}^{2} + 2)} }$

We know,

$\red{ \boxed{ \bf \: \displaystyle\lim_{x \to 0}\bf \: \bigg(1 + x \bigg)^{ \dfrac{1}{x} } = e}}$

So, using this result, we get

$\rm \: = \: \: {e} \: \: ^{\displaystyle\lim_{x \to 0}\sf \: \: \dfrac{ - 4}{ {7x}^{2} + 2} }$

$\rm \: = \: \: {e} \: ^{\dfrac{ - 4}{2} }$

$\rm \: = \: \: {e}^{ - 2}$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\bf \bigg(\dfrac{ {3x}^{2} + 2}{ {7x}^{2} + 2}\bigg)^{\dfrac{1}{ {x}^{2} } } = {e}^{ - 2}$

Alternative Method :-

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \bigg(\dfrac{ {3x}^{2} + 2}{ {7x}^{2} + 2}\bigg)^{\dfrac{1}{ {x}^{2} } }$

Let assume that

$\rm :\longmapsto\:y = \displaystyle\lim_{x \to 0}\rm \bigg(\dfrac{ {3x}^{2} + 2}{ {7x}^{2} + 2}\bigg)^{\dfrac{1}{ {x}^{2} } }$

Taking log on both sides, we get

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0}\rm log \: \bigg(\dfrac{ {3x}^{2} + 2}{ {7x}^{2} + 2}\bigg)^{\dfrac{1}{ {x}^{2} } }$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0}\rm \: {\dfrac{1}{ {x}^{2} } }log \: \bigg(\dfrac{ {3x}^{2} + 2}{ {7x}^{2} + 2}\bigg)$

$\: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log( {x}^{y} ) = y log(x) \bigg \}}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0}\rm \: {\dfrac{log(3 {x}^{2} + 2) - log( {7x}^{2} + 2)}{ {x}^{2} } }$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log( \dfrac{x}{y} ) = logx - logy \bigg \}}$

On applying L - Hospital Rule, we get

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0}\rm \: {\dfrac{ \dfrac{d}{dx} log(3 {x}^{2} + 2) - \dfrac{d}{dx}log( {7x}^{2} + 2)}{ \dfrac{d}{dx}{x}^{2} } }$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0}\rm \: {\dfrac{ \dfrac{1}{ {3x}^{2} + 2} (6x) - \dfrac{1}{ {7x}^{2} + 2}(14x)}{ 2x} }$

$\sf\red{\bigg \{ \because \: \dfrac{d}{dx}logx = \dfrac{1}{x} \bigg \}} \: \: \: and \: \: \red{\bigg \{ \because \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \bigg \}}$

On cancel 2x from numerator and denominator, we get

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0}\rm \: {\dfrac{ \dfrac{1}{ {3x}^{2} + 2} (3) - \dfrac{1}{ {7x}^{2} + 2}(7x)}{ 1} }$

$\rm :\longmapsto\:logy = \dfrac{3}{2} - \dfrac{7}{2}$

$\rm :\longmapsto\:logy = \dfrac{3 - 7}{2}$

$\rm :\longmapsto\:logy = \dfrac{ - 4}{2}$

$\rm :\longmapsto\:logy = - 2$

$\bf\implies \:y = {e }^{ - 2}$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\bf \bigg(\dfrac{ {3x}^{2} + 2}{ {7x}^{2} + 2}\bigg)^{\dfrac{1}{ {x}^{2} } } = {e}^{ - 2}$

Additional Information :-

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

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

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

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

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

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

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

$\red{ \boxed{ \bf \: \displaystyle\lim_{x \to 0}\bf \: \bigg(1 + x \bigg)^{ \dfrac{1}{x} } = e}}$

$\red{ \boxed{ \bf \: \displaystyle\lim_{x \to 0}\bf \: \bigg(1 + kx \bigg)^{ \dfrac{1}{x} } = {e}^{k} }}$