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Mathematics 15 Online
OpenStudy (anonymous):

evaluate lim x π /4 sin2x^tan^2 2x

OpenStudy (nikita2):

as x tends to...?

OpenStudy (dumbcow):

just for clarity, is this what you are asking: \[\lim_{x \rightarrow \pi/4}(\sin 2x)^{\tan^{2} 2x}\]

OpenStudy (anonymous):

\[\lim_{x \rightarrow \pi/4}(\sin2x)^{\tan^2 2x}=\lim_{x \rightarrow \pi/4}(1-(sinx -\cos x)^2)^{\tan^2 2x}\] \[\lim_{x \rightarrow \pi/4}(1-(1-\sin2x))^{\frac{\sin^22x}{(1-\sin2x)(1+2\in2x)}}\]\[(\lim_{x \rightarrow \pi/4}(1-(1-\sin2x))^\frac 1{1-\sin2x})^{\lim_{x \rightarrow \pi/4}\frac {\sin^22x}{1+\sin2x}}\]\[e^{1/2}\]

OpenStudy (anonymous):

hello got it???

OpenStudy (anonymous):

i used the formula..\[\lim_{y \rightarrow 0}(1+y)^{\frac1y}=e\]

OpenStudy (anonymous):

here y=1-sin2x

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