<s>[INST] Trigonometric Limits Question $$\lim _{x\to 0}\frac{\left(\cos \left(\frac{\pi }{2\cos \left(x\right)}\right)\right)}{\sin \left(\sin \left(x\right)^2\right)}=l$$
then find the value of {l}, where {l} denotes fractional part.
I tried multiplying and dividing the denominator with $\sin \left(x\right)^2$ and $x^2$ to simply the denominator, transformed the cos function to sin, repeated the multiplying and dividing once again then I got stuck after that. I reached here, I am not sure how to proceed.
$$\lim _{x\to 0}\frac{\left(\sin \left(\frac{\pi }{2}-\frac{\pi }{2\cos x}\right)\right)}{\frac{\left(\sin \:\left(\sin \:\left(x\right)^2\right)\sin \:\left(x\right)^2x^2\right)}{\sin \left(x\right)^2x^2}}$$
$$=\lim _{x\to 0}\frac{\left(\frac{\sin \left(\frac{\pi }{2}-\frac{\pi }{2\cos x}\right)}{\frac{\pi }{2}-\frac{\pi }{2\cos \left(x\right)}}\right)\left(\frac{\pi \:}{2}-\frac{\pi \:}{2\cos \left(x\right)}\right)}{x^2}$$
$$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{\cos \left(x\right)}\right)}{x^2}$$
$$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(-\frac{1}{2}\right)}{x^2}$$
Could anyone please explain how to proceed with this question and correct any mistakes up to what I solved if there are any?
 [/INST] 
$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{\cos \left(x\right)}\right)}{x^2}=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(-\frac{1}{2}\right)}{x^2}$

This step is incorrect .It should have been:
$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{\cos \left(x\right)}\right)}{x^2}=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(-\frac{1}{2}\right)}{1}$
as $\lim _{x\to 0} \frac{\ cosx-1}{x^2}=-\frac{1}{2} $ and  $\lim _{x\to 0} {\cos x=1}$.
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