Vtyjkyuk

I came across this problem as shown in the title.
Limit of sin(x)sin(1/x) as x approaches 0. I plot the graph using online graphing calculators and found that it is approaching zero. But can anybody please proof it? I am really stuck and don't know where to start.

Also I did try to search the internet and found that the limit of xsin(1/x) equals to zero as x approaches zero. I understand how that work. So my second question is can I say that limit of
sin(x)sin(1/x) is the same as limit xsin(1/x) when x approaches zero, since limit of sin(x)/x equals 1 when x approaches zero?

HINT

Let observe that since $forall theta: , |sin theta|le 1$

$$0le |sin x cdot sin(1/x)|=|sin x| cdot |sin(1/x)|le |sin x|$$

then refer to squeeze theorem.

Note that the result you are referring to, that is

$$xcdot sinleft(frac1xright)to 0$$

can be obtained in the same way by squeeze theorem

$$0le |x cdot sin(1/x)|=|x| cdot |sin(1/x)|le |x|to 0$$

once we know that, we can also proceed by standards limit and conclude that

$$sin x cdot sinleft(frac1xright)=fracsin xx cdot x cdot sinleft(frac1xright)to 1 cdot 0 = 0$$

but it is a pretty convolute way since we can apply directly the squeeze theorem to the given limit.

HINT:
$$
|sin xsin(1/x)|leq|x|cdot|sin(1/x)|
$$