Question

Let $r={lim}_{x\to 0}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x}};m={lim}_{x\to 0}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}$, then $r+m=$

Answer 1

We have,

$r=\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x}}\left(1^{\infty }form\right)$

$r=e^{\underset{x\to 0}{lim}\frac{1}{x}(\frac{\sin x}{x}-1)}$

$=e^{\underset{x\to 0}{lim}\frac{\sin x-x}{x^2}}$

$e^{\underset{x\to 0}{lim}\frac{\cos x-1}{2x}}$

$e^{\underset{x\to 0}{lim}\frac{-\sin x}{2}}$

$r=e^0$

$\therefore r=1$

Since,

$m=\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}\left(1^{\infty }form\right)$

$e^{\underset{x\to 0}{lim}\frac{1}{x^2}\left(\frac{\sin x-x}{x}\right)}$

$e^{\underset{x\to 0}{lim}\left(\frac{\sin x-x}{x^3}\right)}$

$e^{\underset{x\to 0}{lim}\left(\frac{\cos x-1}{3x^2}\right)}$

$e^{\underset{x\to 0}{lim}\left(\frac{-\sin x-0}{6x}\right)}$

$e^{\underset{x\to 0}{lim}\left(\frac{-\cos x-0}{6}\right)}$

$m=e^{-\frac{1}{6}}$

$\therefore r+m=1+e^{-\frac{1}{6}}$

Hence, this is the answer.