Calculation of $\lim_{x\rightarrow 0}\frac{\sin (\pi\cos^2 x)}{x^2}$

Asked Apr 08 '14 at 05:41

Active Apr 08 '14 at 05:56

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Calculation of $\displaystyle \lim_{x\rightarrow 0}\frac{\sin (\pi\cos^2 x)}{x^2}$

$\bf{My\; Try::}$ Given $\displaystyle \lim_{x\rightarrow 0}\frac{\sin (\pi\cos^2 x)}{x^2} = \lim_{x\rightarrow 0}\frac{\sin (\pi (1-\sin^2 x))}{x^2}$

$\displaystyle \lim_{x\rightarrow 0}\frac{\sin (\pi \sin^2 x)}{\pi\sin^2 x} \times \pi \times \lim_{x\rightarrow 0} \frac{\sin^2 x}{x^2} = \pi$

Is there is any other method by which we can solve the above question.

If yes, The please help me .

Thanks

asked Apr 08 '14 at 05:41

juantheron

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3

Are you familiar with L'Hospital's Rule? But your method is nicer. – André Nicolas Apr 08 '14 at 05:45
I don't sse what the second line has to do with the first line in your "My Try" section. Where did the $1-$ go? – Thomas Andrews Apr 08 '14 at 05:49
$\sin(\pi -t)=\sin t$. – André Nicolas Apr 08 '14 at 05:50
if we do L'Hospital rule we may get answer as zero..am i right? – santosh kambhampati Apr 08 '14 at 05:56
If we use L'Hospital's Rule, we get $\pi$. – André Nicolas Apr 08 '14 at 06:01
hm yaa u r right i tried it again thank u :) – santosh kambhampati Apr 08 '14 at 06:05
L'Hopital is very messy after the first round of derivative. Is there any other method? – MangoPizza Aug 07 '22 at 18:40

Calculation of $\lim_{x\rightarrow 0}\frac{\sin (\pi\cos^2 x)}{x^2}$

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