What is the limit of sin (1/x) when x tends to zero?

Limit sin(1/x) when x tends to 0 is not defined

can be proved simply by multiplying and dividing by x then xsin(1/x)/x becomes 1/x as xsin(1/x)or for that matter sin(1/x)/1/x = 1 and limit reduces to 1/x which doesnt exist 

also the proof can be that when x approcashes 0 from positive side 1/x tends to positive infinty

and limit (right0 becomes sin(infinity)

but when from left side 1/x tends to negative infinty so limit becomes -sin(infinit) which both can never b equal 

so limit doesnt exist

dude,this is exactly what i assumed,,,,actually u can check it outin objective rd sharma..but thre was no option for our answer, and solution said that the value will oscillate between 1 and -1.. u have any idea about that?

As x tends to zero the value of 1/x tends to either ∞ or -∞. In this situation we will not be sure about the exact value of 1/x. As a result we will not be sure about the exact/approaching value of sin(1/x). We cant say anything about the value of sine function untill we know the angle and in this question we are not sure about the angle as at infinity it can take any value. We will be sure that the value of sin(1/x) will lie in [-1, 1] but not sure about a unique value. As in limits, it exists only when we get a unique value. Hence we will say that the limit does not exist.