Math / CalculusQuestion6. Evaluate limx→π2(sinx−cosx)tanx\lim _{x \rightarrow \frac{\pi}{2}}(\sin x-\cos x)^{\tan x}limx→2π(sinx−cosx)tanx [5 pts]Studdy SolutionSince limx→π2lny=0\lim_{x \to \frac{\pi}{2}} \ln y = 0limx→2πlny=0, we have:limx→π2y=e0=1 \lim_{x \to \frac{\pi}{2}} y = e^0 = 1 x→2πlimy=e0=1The value of the limit is:1 \boxed{1} 1View Full Solution - FreeWas this helpful?