Lời giải:
Ta có:
$lim_{x->\frac{\pi}{2}}(tgx)^{2x-\pi}=e^{lim_{x->\frac{\pi}{2}}(2x-\pi).lntgx}$
$lim_{x->\frac{\pi}{2}}(2x-\pi)lntgx=lim_{x->\frac{\pi}{2}}\frac{(lntgx)'}{(\frac{1}{2x-\pi})'}$
$=-lim_{x->\frac{\pi}{2}}\frac{\frac{1}{tgx}.\frac{1}{cos^2x}}{\frac{2}{(2x-\pi)^2}}=-\frac{1}{2}lim_{x->\frac{\pi}{2}}\frac{(2x-\pi)^2}{sinx.cosx}=-\frac{1}{2}lim_{x->\frac{\pi}{2}}\frac{(2x-\pi)^2}{cosx}=-\frac{1}{2}lim_{x->\frac{\pi}{2}}\frac{[(2x-\pi)^2]'}{(cosx)'}=-\frac{1}{2}lim_{x->\frac{\pi}{2}}\frac{4.(2x-\pi)}{-sinx}=0$
=>$lim_{x->\frac{\pi}{2}}(tgx)^{2x-\pi}=e^0=1$
