Question

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$\displaystyle\lim_{x \to 2}\ \bf{( \frac{tanx - sinx}{ {sin}^{3}x } )}$



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Answer 1

Appropriate Question :-

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{tanx - sinx}{ {sin}^{3} x}$

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{tanx - sinx}{ {sin}^{3} x}$

If we substitute directly x = 0,

$\rm \:  =  \: \dfrac{tan0 - sin0}{ {sin}^{3} 0}$

$\rm \:  =  \: \dfrac{0 - 0}{0}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{tanx - sinx}{ {sin}^{3} x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{ \dfrac{sinx}{cosx} - sinx }{ {sin}^{3} x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \dfrac{sinx\bigg[\dfrac{1}{cosx} - 1 \bigg]}{ {sin}^{3} x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \dfrac{1 - cosx}{cosx \: {sin}^{2} x}$

$\rm \:  =  \:\displaystyle\lim_{x \to 0}\rm \frac{1}{cosx} \times \displaystyle\lim_{x \to 0}\rm \dfrac{1 - cosx}{1 - \: {cos}^{2} x}$

$\rm \:  =  \: 1 \times \displaystyle\lim_{x \to 0}\rm \frac{1 - cosx}{(1 - cosx)(1 + cosx)}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{1}{1 + cosx}$

$\rm \:  =  \: \dfrac{1}{1 + cos0}$

$\rm \:  =  \: \dfrac{1}{1 + 1}$

$\rm \:  =  \: \dfrac{1}{2}$

Hence,

$\purple{\rm\implies \:\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx - sinx}{ {sin}^{3} x} = \frac{1}{2}}}}$

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MORE TO LEARN

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

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

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

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

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

Answer 2

Answer :

$\displaystyle\lim_{x \to 2}\ \bf{( \frac{tanx - sinx}{ {sin}^{3}x } )} = \frac{1}{2}$

Explanation :

Transform the function in this way:

$\\ \tt : \implies \frac{tanx-sinx}{x^3 }= \frac{1}{x^3}( \frac {sinx}{cosx} - sinx)$

$\\ \tt : \implies \frac{tanx - \: sinx}{ {x}^{3} } = \frac{1}{ {x}^{3} } (\frac{sinx - sinxcosx}{cosx} )$

$\\ \tt : \implies \frac{tanx - \: sinx}{ {x}^{3} } = \frac{sinx}{ {x}^{3} } (\frac{1 - cosx}{cosx} )$

$\\ \tt : \implies \frac{tanx - \: sinx}{ {x}^{3} }$

$\\ \tt : \implies \: ( \frac{sinx}{x} ) ( \frac{1 - cosx}{ {x}^{2} } ) (\frac{1}{cosx} )$

We can use now the well known trigonometric limit:

$\\ : \implies \tt \: \lim_{x \to 2}\ \bf{\frac{sinx}{ x } } = 1$

and using the trigonometric identity:

$\\ : \implies \tt \: sin^2 \alpha = \frac{ (1-cos2alpha)}{2}$

we have:

$\\ : \implies\tt\lim_{x \to 2} {( \frac{1 - cosx}{ x {}^{2} } )} =\lim_{x \to 2} \frac{2sin {}^{2}( \frac{x}{2} ) }{ {x}^{2} } = \frac{1}{2}$

$\\ : \implies\tt\lim_{x \to 2}( \frac{sin( \frac{ x}{2} )}{ \frac{x}{2} } ) {}^{2} = \frac{1}{2}$

While the third function is continuous so:

$\\ : \implies\tt\lim_{x \to 2} \: \: \: \frac{1}{cosx} = \frac{1}{1} = 1$

and we can conclude that:

$: \implies\tt\lim_{x \to 2}{( \frac{tanx - sinx}{ x^{3} } )}$

$: \implies\tt\lim_{x \to 2} \: ( \frac{sinx}{x} ) ( \frac{1 - cosx}{ {x}^{2} } ) (\frac{1}{cosx} )$

$: \implies \tt \: 1 \times \frac{1}{2} \times 1 = \frac{1}{2}$