Use The Most Standard Method!

For $\lim_{x\to \infty}\left(\frac{x}{\sin(x)}\right)$ Use the Limit Divergence Criterion test.

If there exist two sequences, $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ with

$\begin{aligned}x_n \not=c \space\ce{and}\space y_n \not =c \\ \lim (x_n)=\lim (y_n) c \\ \lim f(x_n)\not = \lim f(y_n) \\ \end{aligned}$

Then $\lim_{x\to c}f(x) = \infty$ .

$\LARGE \text{FIN!!!}$

We have to evaluate: $\displaystyle \lim_{x \rightarrow \infty} \frac{x}{\sin x} = \frac{1}{lim_{x \rightarrow \infty} \frac{\sin x}{x}}$

Let $\displaystyle y=\frac{1}{x}$ then $\displaystyle y \rightarrow 0$ . Now, we have

$\displaystyle \frac{1}{\lim_{y \rightarrow 0} \frac{\sin \frac{1}{y}}{\frac{1}{y}}}$

$= \displaystyle \frac{1}{\lim_{y \rightarrow 0} y \sin \frac{1}{y}}$

$= \displaystyle \frac{1}{0 \times \sin \infty}$

$= \displaystyle \frac{1}{0} \rightarrow$ Limit does not exist

Since the value of $\sin{x}$ lies between 1 and -1 and keeps moving, so the limit does not exist.