$\lim_{x\to\infty}(\frac{\sin(x)}{x})^{\frac{1}{x}}$.

$\lim_{x\to\infty}(\frac{\sin(x)}{x})^{\frac{1}{x}}$ is equal to

$1.$ $0$.

$2$. $1$.

$3$. $-1$.

$4$. None .

Answer is given to be $1$ but my doubt is if I try by sequential criteria by taking sequence $<n\pi>$ which is divergent to $\infty$ and $<f(a_n)>\to 0$. So limit can’t be $1$. Where is mistake ? Please reply . Thank you.

Solution 1:

I got 4. None. On the one hand: For each integer $n$, pick $x_n$ close enough to $2\pi n$ such that the inequality $\frac{\sin x}{x} = e^{-n^2}$ holds i.e., $\frac{\sin x}{x}$ is extremely close to $0$ while $x$ isn't too large.

$$\lim_{n \rightarrow \infty} \left(\frac{\sin x_n}{x_n}\right)^{\frac{1}{x_n}} = \lim_{n \rightarrow \infty} (e^{-n^2})^{\frac{1}{x_n}} = e^{-\Omega(n)} = 0.$$

On the other hand, pick $x'_n$ such that $x'_n=2\pi n +\frac{\pi}{2}$. Then one can check

$$\lim_{n \rightarrow \infty} \left(\frac{\sin x'_n}{x'_n}\right)^{\frac{1}{x'_n}} = \lim_{n \rightarrow \infty} \left(\theta \left(\frac{1}{n}\right)\right)^{\theta\left(\frac{1}{n}\right)}$$ $$= \lim_{n \rightarrow \infty} e^{-O\left(\frac{\ln n}{n} \right)} = 1.$$

ETA: I noted in the comments below this answer but I might as well edit here: My argument could infact be made even simpler: Pick $x_n$ as above to be $x_n=2\pi n$. Then $(\frac{\sin x_n}{x_n})^{\frac{1}{x_n}}=0$ for each positive $n$.