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Question: f(x)=\(\lim_{n \rightarrow \infty}\){sinx + 2sin<sup>2</sup>x + 3sin<sup>3</sup>x +...+ n sin<sup>n<...

f(x)=lim⁡n→∞\lim_{n \rightarrow \infty}{sinx + 2sin2x + 3sin3x +...+ n sinnx}. If x ¹ np + π2\frac{\pi}{2}, n Ī I, then lim⁡x→π/2[(1–sin⁡x)2f(x)]1sin⁡x–1\lim_{x \rightarrow \pi/2}\lbrack(1–\sin x)^{2}f(x)\rbrack^{\frac{1}{\sin x–1}}is equal to :

A

1

B

0

C

e

D

e2

Answer

e

Explanation

Solution

f(x) = sin⁡x(1–sin⁡x)2\frac{\sin x}{(1–\sin x)^{2}}

Now lim⁡x→π/2\lim_{x \rightarrow \pi/2} [(1–sin⁡x)2f(x)]1sin⁡x–1\lbrack(1–\sin x)^{2}f(x)\rbrack^{\frac{1}{\sin x–1}}

= lim⁡x→π/2(sin⁡x)1sin⁡x–1\lim_{x \rightarrow \pi/2}(\sin x)^{\frac{1}{\sin x–1}}

= elim⁡x→π/2sin⁡x–1sin⁡x–1e^{\lim_{x \rightarrow \pi/2}\frac{\sin x–1}{\sin x–1}} = e