Question

Find the given limit :
$\displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi \: cos^{2} (x) )}{ {x}^{2} } }$
​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi \: cos^{2} (x) )}{ {x}^{2} } }$

If we substitute directly x = 0, we get

$\rm \: = \: \dfrac{ sin(\pi \: cos^{2} (0) )}{ {0}^{2} }$

$\rm \: = \: \dfrac{ sin(\pi \: \times 1)}{ 0}$

$\rm \: = \: \dfrac{ sin(\pi)}{ 0}$

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

which is indeterminant form

So,

$\rm :\longmapsto\:\displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi \: cos^{2} (x) )}{ {x}^{2} } }$

can be rewritten as

$\rm \: = \: \displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi[1 - sin^{2}x] )}{ {x}^{2} } }$

$\rm \: = \: \displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi - \pi \: sin^{2}x )}{ {x}^{2} } }$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \: sin(\pi - x) = sinx \: }}}$

So, using this we get

$\rm \: = \: \displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi \: sin^{2}x )}{ {x}^{2} } }$

$\rm \: = \: \displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi \: sin^{2}x )}{\pi {sin}^{2}x} } \times \dfrac{\pi \: {sin}^{2}x}{ {x}^{2} }$

$\rm \: = \pi\: \displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi \: sin^{2}x )}{\pi {sin}^{2}x} } \times \displaystyle\lim_{x \to 0}\tt \dfrac{\: {sin}^{2}x}{ {x}^{2} }$

We know,

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

$\red{\sf[As \: x \to 0\sf\implies sinx \to 0\sf\implies {sin}^{2}x \to 0\sf\implies \pi{sin}^{2}x \to0]}$

So, using this we get

$\rm \: = \: \pi \times 1 \times 1$

$\rm \: = \: \pi$

Hence,

$\sf\implies \: \boxed{\tt{ \: \: \displaystyle \tt{ \lim_{x \to0} \: \dfrac{ sin(\pi \: cos^{2} (x) )}{ {x}^{2} } } = \pi \: \: }}$

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EXPLORE MORE

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

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

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

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

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

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

Answer 2

$\displaystyle \sf \red{ \lim_{x \to 0} \frac{\sin{\left(\pi \cos^{2}{\left(x \right)} \right)}}{x^{2}}}$

$\displaystyle \sf {\color{red}{\lim_{x \to 0} \frac{\sin{\left(\pi \cos^{2}{\left(x \right)} \right)}}{x^{2}}} = \color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\sin{\left(\pi \cos^{2}{\left(x \right)} \right)}\right)}{\frac{d}{dx}\left(x^{2}\right)}}}$

$\displaystyle \sf \red{\color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\sin{\left(\pi \cos^{2}{\left(x \right)} \right)}\right)}{\frac{d}{dx}\left(x^{2}\right)}} = \color{red}{\lim_{x \to 0}\left(- \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)} \cos{\left(\pi \cos^{2}{\left(x \right)} \right)}}{x}\right)}}$

$\displaystyle \sf \red{\color{red}{\lim_{x \to 0}\left(- \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)} \cos{\left(\pi \cos^{2}{\left(x \right)} \right)}}{x}\right)} = \color{red}{\lim_{x \to 0} \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)}}{x}}}$

$\displaystyle \sf{\color{red}{\lim_{x \to 0} \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)}}{x}} = \color{red}{\pi \lim_{x \to 0} \frac{\sin{\left(x \right)} \cos{\left(x \right)} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)}}{x}}}$

$\displaystyle \sf \red{\pi \color{red}{\lim_{x \to 0} \frac{\sin{\left(x \right)} \cos{\left(x \right)} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)}}{x}} = \pi \color{red}{\lim_{x \to 0} \frac{\sin{\left(x \right)}}{x} \lim_{x \to 0} \cos{\left(x \right)} \lim_{x \to 0} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)}}}$

$\displaystyle \sf \red{\pi \lim_{x \to 0} \frac{\sin{\left(x \right)}}{x} \lim_{x \to 0} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)} \color{red}{\lim_{x \to 0} \cos{\left(x \right)}} = \pi \lim_{x \to 0} \frac{\sin{\left(x \right)}}{x} \lim_{x \to 0} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)} \color{red}{1}}$

${\displaystyle \sf\red{\pi \lim_{x \to 0} \frac{\sin{\left(x \right)}}{x} \color{red}{\lim_{x \to 0} \cos{\left(\pi \sin^{2}{\left(x \right)} \right)}} = \pi \lim_{x \to 0} \frac{\sin{\left(x \right)}}{x} \color{red}{1}}}$

$\displaystyle \sf \red{\pi \color{red}{\lim_{x \to 0} \frac{\sin{\left(x \right)}}{x}} = \pi \color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\sin{\left(x \right)}\right)}{\frac{d}{dx}\left(x\right)}}}$

$\displaystyle \sf \red{\pi \color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\sin{\left(x \right)}\right)}{\frac{d}{dx}\left(x\right)}} = \pi \color{red}{\lim_{x \to 0} \cos{\left(x \right)}}}$

$\displaystyle \sf \red{\pi \color{red}{\lim_{x \to 0} \cos{\left(x \right)}} = \pi \color{red}{1}}$

Therefore,

$\displaystyle \sf \red{\lim_{x \to 0} \frac{\sin{\left(\pi \cos^{2}{\left(x \right)} \right)}}{x^{2}} = \pi}$