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Question: If f(n) = \(\operatorname { Lim } _ { x \rightarrow 0 }\) \(\left( \left( 1 + \sin\frac{x}{2} \right...

If f(n) = Lim⁡x→0\operatorname { Lim } _ { x \rightarrow 0 } ((1+sin⁡x2)(1+sin⁡x22).....(1+sin⁡x2n))1x\left( \left( 1 + \sin\frac{x}{2} \right)\left( 1 + \sin\frac{x}{2^{2}} \right).....\left( 1 + \sin\frac{x}{2^{n}} \right) \right)^{\frac{1}{x}} thenLimn→∞\underset{n \rightarrow \infty}{Lim}f(n) =

A

1

B

e

C

0

D

∞

Answer

e

Explanation

Solution

f(n) = Limx→0e1x((1+sin⁡x2)(1+sin⁡x22)....(1+sin⁡x2n)−1)\underset{x \rightarrow 0}{Lim}e^{\frac{1}{x}\left( \left( 1 + \sin\frac{x}{2} \right)\left( 1 + \sin\frac{x}{2^{2}} \right)....\left( 1 + \sin\frac{x}{2^{n}} \right) - 1 \right)}

= Limx→0e(1+(sin⁡x2+sin⁡x22+......+sin⁡x2n)+(sin⁡x2sin⁡x22+......)+.....−1)x\underset{x \rightarrow 0}{Lim}e^{\frac{\left( 1 + \left( \sin\frac{x}{2} + \sin\frac{x}{2^{2}} + ...... + \sin\frac{x}{2^{n}} \right) + \left( \sin\frac{x}{2}\sin\frac{x}{2^{2}} + ...... \right) + ..... - 1 \right)}{x}}

= Limx→0e(sin⁡x2x+sin⁡(x22)x+.......+sin⁡(x2n)x)\underset{x \rightarrow 0}{Lim}e^{\left( \frac{\sin\frac{x}{2}}{x} + \frac{\sin\left( \frac{x}{2^{2}} \right)}{x} + ....... + \sin\frac{\left( \frac{x}{2^{n}} \right)}{x} \right)}

= e(12+122+.....12n)e^{\left( \frac{1}{2} + \frac{1}{2^{2}} + .....\frac{1}{2^{n}} \right)}

Sum of infinite G.P. = e1/21−12e^{\frac{1/2}{1 - \frac{1}{2}}}= e