Limit

For $\lim_{x \to \infty} \left(\frac{\sin(x)}{x}\right)$ apply Squeeze Theorem

$^{-}1 \le \sin(x) \le 1$

$\lim_{x \to \infty}\left(-\frac{1}{x}\right) \le \lim_{x \to \infty}\left(\frac{\sin(x)}{x}\right) \le \lim_{x \to \infty}\left(\frac{1}{x}\right)$

$\lim_{x \to \infty}\left(-\frac{1}{x}\right) = 0$

$\lim_{x \to \infty}\left(\frac{1}{x}\right) = 0$

$\therefore \space \space \lim_{x \to \infty} \left(\frac{\sin(x)}{x}\right)$ is $0$ . $\square$

$\LARGE \text{ADIOS!!!}$

Or it can be done like this too.. Put x=1/y. Now as x tends to infinity ,y tends to 0.lim sin(1/y)÷(1/y) Or,lim y.sin(1/y);y tends to 0 Or, 0×sin(1/y) =0

The Range of sinx is [-1,1] .i.e. it will be a finite value. In denominator it is infinite. Therfore, finite/infinite value will tend to zero.

$\large I = \lim_{x\to\infty} \frac{sinx}{x}$

Apply L-Hospital's rule ,

$\large I = \lim_{x\to\infty}\frac{\frac{d}{dx}(sinx)}{\frac{d}{dx}(x)}\ = \lim_{x\to\infty} cosx = cos(\infty) = 0$