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We can use the sandwich theorem. We know that: $-1\le \sin { \left( x \right) } \le 1$

Thus, $\lim _{ x\rightarrow \infty }{ \frac { -1 }{ x } } \le \lim _{ x\rightarrow \infty }{ \frac { \sin { \left( x \right) } }{ x } } \le \lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } }$

Hence, $0\le \lim _{ x\rightarrow \infty }{ \frac { \sin { \left( x \right) } }{ x } } \le 0$

The answer is 1..(obviously not since x $\rightarrow \infty$ (not 0) ). Notice that as x $\rightarrow \infty$ , 1/x tends to zero while sin x oscillates between 1 and -1. Hence, $\large \lim_{x\rightarrow \infty} \frac{1}{x} \times \sin x$ $=0 \times \text{ (a finite number )}$ $~~~~~~~~~~~~~~~~~~\Large=\boxed 0$

The limit that doesn't actually exist is sin (1/x). Sin x doesn't go further from -1,1 when x grows indefinitely, and the denominator grows indefinitely, which means that the number is reaching 0. Solution: 0.

As the numerator, sin(x), approaches infinity, it always stays within -1 and 1. But as the denominator, x, approaches infinity, it gets larger and larger. The numerator gets divided by larger and larger numbers, resulting in it constantly growing smaller (when the top is positive) or larger (when the topis negative) and consequently reaching zero

L'hopital's rule. The limit = 0.

sinx is always b/w -1 and 1 so it is a finite number. also when we apply limits,x=(infinity)=1/0 which will give the solution a finite number*0=0 hence the answer

Actually the limit doesn't exist