Math  /  Calculus

Question6. Evaluate limxπ2(sinxcosx)tanx\lim _{x \rightarrow \frac{\pi}{2}}(\sin x-\cos x)^{\tan x} [5 pts]

Studdy Solution
Since limxπ2lny=0\lim_{x \to \frac{\pi}{2}} \ln y = 0, we have:
limxπ2y=e0=1 \lim_{x \to \frac{\pi}{2}} y = e^0 = 1
The value of the limit is:
1 \boxed{1}

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