Tutor profile: Igor B.

Proof $\sin \left( \frac{\pi}{2}+\alpha \right) =\cos \alpha $

$\sin \left( \frac{\pi}{2}+\alpha \right) =\sin \frac{\pi}{2}\cos \alpha +\cos \frac{\pi}{2}\sin \alpha =1\cdot \cos \alpha +0\cdot \sin \alpha =\cos $

Evaluate $\underset{n\rightarrow \infty}{lim}\frac{\sqrt{x+1}-1}{x}=?$

$\lim_{x\rightarrow 0} \frac{\sqrt{x+1}-1}{x}=\lim_{x\rightarrow 0} \frac{\sqrt{x+1}-1}{x}\cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\lim_{x\rightarrow 0} \frac{\left( \sqrt{x+1} \right) ^2-1}{x\left( \sqrt{x+1}+1 \right)}=\lim_{x\rightarrow 0} \frac{x+1-1}{x\left( \sqrt{x+1}+1 \right)}=\lim_{x\rightarrow 0} \frac{\cancel{x}}{\cancel{x}\left( \sqrt{x+1}+1 \right)}=\lim_{x\rightarrow 0} \frac{1}{\sqrt{x+1}+1}=\frac{1}{2}$

Evaluate $\log _23$ ?

$x=\log _23\Rightarrow 3=2^x\\\left( 2^x \right) ^3=\left( 3 \right) ^3=27<32=2^5\\2^{3x}<2^5\\3x<5\\x<\frac{5}{3}\\\left( 2^x \right) ^2=\left( 3 \right) ^2=9>8=2^3\\2^3<2^{2x}\\3<2x\\x>\frac{3}{2}\\\frac{3}{2}<x<\frac{5}{3}$