A problematic limit

Calculus Level 4

$\large \displaystyle\lim _{ x\to \infty } x^{ 2 }\sin {\left( \ln { \sqrt{\cos{\dfrac{\pi}{x}} }} \right) }= \, ?$

-\frac{\pi ^2}{8}

None of these choices

0

-\frac{\pi ^2}{2}

-\frac{\pi ^2}{4}

2 solutions

Swagat Panda
Aug 21, 2016

What happened with function Sin in the third step?

Catalina Sánchez - 4 years, 9 months ago

It's converted into sin(x)/x form where x tends to zero

Kshitij Ghode - 4 years, 9 months ago

exactly. :)

Swagat Panda - 4 years, 9 months ago

$\displaystyle \lim_{\theta \rightarrow 0} {\frac{\sin{\theta}}{\theta}}=1$

Swagat Panda - 4 years, 9 months ago

Prakhar Bindal
Aug 21, 2016

Firstly let x = 1/t

therefore limit becomes t approaching zero

Term inside sin is tending to zero hence we use sinx = x

So finally limit is ln(cos(pi*t))/t^2

Apply L Hospital rule once and then write tanx = x to get required answer

Pretty much the same, except for the substitution and the L'Hopital's in the penultimate step. Short and sweet. :)

Swagat Panda - 4 years, 9 months ago

I see from your problem solving skills that you will get a very good AIR in the JEE exam.Keep it up.

Indraneel Mukhopadhyaya - 4 years, 9 months ago

Thanks a lot bhaiya! hope you understand hindi

Where are you studying currently?

Prakhar Bindal - 4 years, 9 months ago

Indraneel Mukhopadhyaya - 4 years, 9 months ago

nice prakhar short and sweet

Pawan pal - 4 years, 9 months ago

Thanks bro!

Prakhar Bindal - 4 years, 9 months ago