Question

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$lim_{x to 1} frac{sin^2(pi2^x)}{ln(cos(pi2^x))}$

Answer 1

$displaystyle lim_{x to 1} frac{sin^2(pi2^x)}{lncos(pi2^x)}=lim_{x to 1} frac{sin^2(pi2^x)}{lnsqrt{1-sin^2(pi2^x)}}=left{begin{array}{ccc}sin^2(pi 2^x)=t\ tto0end{array}right}=lim_{t to 0}frac{t}{0.5ln(1-t)}=\ \ \ =lim_{t to 0}frac{t}{-frac{t}{2}+o(t)}=-2$