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Question: \(\lim_{m \rightarrow \infty}\left( \cos\frac{x}{m} \right)^{m}\)=0...

lim⁡m→∞(cos⁡xm)m\lim_{m \rightarrow \infty}\left( \cos\frac{x}{m} \right)^{m}=0

A

0

B

e

C

1/e

D

1

Answer

1

Explanation

Solution

lim⁡m→∞(cos⁡xm)m=lim⁡m→∞[1+(cos⁡xm−1)]m\lim_{m \rightarrow \infty}\left( \cos\frac{x}{m} \right)^{m} = \lim_{m \rightarrow \infty}\left\lbrack 1 + \left( \cos\frac{x}{m} - 1 \right) \right\rbrack^{m}

=lim⁡m→∞[1−(−cos⁡xm+1)]m= \lim_{m \rightarrow \infty}\left\lbrack 1 - \left( - \cos\frac{x}{m} + 1 \right) \right\rbrack^{m}

=lim⁡m→∞[1−2sin⁡2x2m]m= \lim_{m \rightarrow \infty}\left\lbrack 1 - 2\sin^{2}\frac{x}{2m} \right\rbrack^{m}

=elim⁡m→∞−(2sin⁡2x2m)m=elim⁡m→∞−2(sin⁡x2mx2m)2(x24m2)m=e−2lim⁡m→∞x24m=e0=1= e^{\lim_{m \rightarrow \infty} - \left( 2\sin^{2}\frac{x}{2m} \right)m} = e^{\lim_{m \rightarrow \infty} - 2\left( \frac{\sin\frac{x}{2m}}{\frac{x}{2m}} \right)^{2}\left( \frac{x^{2}}{4m^{2}} \right)m} = e^{- 2\lim_{m \rightarrow \infty}\frac{x^{2}}{4m}} = e^{0} = 1.