Question

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

Answer 1

$\displaystyle \lim_{x \to 1} \frac{\sin^2(\pi2^x)}{\ln\cos(\pi2^x)}=\lim_{x \to 1} \frac{\sin^2(\pi2^x)}{\ln\sqrt{1-\sin^2(\pi2^x)}}=\left\{\begin{array}{ccc}\sin^2(\pi 2^x)=t\\ t\to0\end{array}\right\}=\lim_{t \to 0}\frac{t}{0.5\ln(1-t)}=\\ \\ \\ =\lim_{t \to 0}\frac{t}{-\frac{t}{2}+o(t)}=-2$