Evaluate $\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x}$ .

Hint: For solving this question we just need to know one thing that is the interval of the function. As the interval of the $\sin $function varies between $0\& 1$ . Therefore by using this we are now able to answer this question.

Complete step-by-step answer:
As we know that the $\sin x$ oscillates between $0\& 1$ and hence we can say that at any value of $x$ the interval will be as $ - 1 < = \sin x < = 1$ .
So from this, we can say that dividing on dividing anything by infinity, will give us infinitesimal. Since $x$ tends to the infinity and that is we can say it tends to $0$ .
As we can see the intervals. So no matter what the input, $\sin x$ will just oscillate between $0\& 1$ . Since the numerator stays relatively the same and the denominator will get blown up, so the $\frac{{\sin x}}{x}$ will become infinitesimally small and hence it will approach zero.
Since the deviation of the value is negligible, therefore from this it will be concluded that the answer will be equivalent to $0$ .
Hence, the answer will be $0$ .

Note: So for solving this type of question, we need to remember the limits and the intervals for it. For this practice, a lot of questions will help us to remember the intervals. You simply need to know the overall term in every arrangement's expansion. For instance, $\sin $ the arrangement has all odd power terms while $\cos $ has even power terms. On the off chance that you can't recollect the overall term, recall every arrangement till the third term just as a rule they won't ask past this.