Question

${lim}_{x\to \frac{\pi }{2}}\frac{2^{-\cos x}-1}{x(x-\frac{\pi }{2})}$

Answer 1

${lim}_{x\to \frac{\pi }{2}}\frac{2^{-\cos x}-1}{x(x-\frac{\pi }{2})}Putx=\frac{\pi }{2}+hash=x-\frac{\pi }{2}asx\to \frac{\pi }{2},h\to 0={lim}_{h\to 0}\frac{2^{-\cos (\frac{\pi }{2}+4)-1}}{(\frac{\pi }{2}+4)\left(h\right)}={lim}_{h\to 0}\frac{2^{\sin h}-1}{sinh}.\frac{\sin h}{h}.\frac{1}{(\frac{\pi }{2}+h)}={lim}_{h\to 0}\frac{2^{sinh}-1}{\sin h}\times {lim}_{h\to 0}\frac{\sin h}{h}{lim}_{h\to 0}\frac{1}{(\frac{\pi }{2}+h)}={\log }_e^2\left(1\right)\times \frac{2}{\pi }=\frac{2}{\pi }{\log }_e^2$