# Limits of sin(x)sin(1/x) when x approaches 0

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?

• Observe that $|\sin(1/x)| \leq 1$, so $|\sin(x)\sin(1/x)| \leq |\sin(x)|$. – Bungo Sep 4 '18 at 7:16
• thanks a lot everyone – JoisBack Sep 4 '18 at 7:25
• @JoisBack Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE – user Oct 23 '18 at 20:59

HINT

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

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

then refer to squeeze theorem.

Note that the result you are referring to, that is

$$x\cdot \sin\left(\frac1x\right)\to 0$$

can be obtained in the same way by squeeze theorem

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

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

$$\sin x \cdot \sin\left(\frac1x\right)=\frac{\sin x}x \cdot x \cdot \sin\left(\frac1x\right)\to 1 \cdot 0 = 0$$

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

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