Question

If ${lim}_{x\to 0}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}\pi }=z$, where $i=\sqrt{-1}$, then $\text{Re}\left(z^3\right)$ will be

• A. $-1$
• B. Zero
• C. $\pi$
• D. $1$

Answer 1

The correct option is B Zero

Let $z={lim}_{x\to 0}{\left\{{(1+\frac{\sin x-x}{x})}^{\frac{x}{\sin x-x}}\right\}}^{\frac{\sin x-x}{x^3}\times i\pi }=e^{-i\frac{\pi }{6}}$
$\text{Arg}\left(z^3\right)=-\frac{\pi }{2}$