Question

$\displaystyle\lim _{n \rightarrow \infty}\left(\cos \frac{x}{n}\right)^{n}$ is equal to

• A. $e^{1}$
• B. $e^{-1}$
• C. $1$
• D. None of these

Answer 1

Answer: (C)

$\displaystyle\lim _{n \rightarrow \infty}\left(\cos \frac{x}{n}\right)^{n}=e^{\displaystyle\lim _{n \rightarrow \infty} n\left(\cos \frac{x}{n}-1\right)}$
$=e^{\displaystyle\lim _{n \rightarrow \infty}-n \cdot 2 \sin ^{2}\left(\frac{x}{2 n}\right)}$
$= e ^{-2 \displaystyle\lim _{n \rightarrow \infty}\left(\frac{\sin \left(\frac{x}{2 n}\right)}{\frac{x}{2 n}}\right)^{2} \cdot \frac{x^{2}}{4 n^{2}} \cdot n}$
$=e^{-2 \times \displaystyle\lim _{n \rightarrow \infty} \frac{x^{2}}{4 n}}=e^{0}=1$