What is $\lim\limits_{x\to 0}x^2\cos(2/x)$?

Question

I have to evaluate $$\lim_{x\to 0}x^2\cos(2/x)$$ using one or more of the limit laws.

I am using the multiplication law and I am wondering if I am on the right track here?

I have split it up to: $$\left(\lim_{x\to 0}x^2\right)\left(\lim_{x\to 0}\;\cos(2/x)\right)$$ Since $\lim\limits_{x\to 0}x^2 = 0$, is the final answer $0$?

Thanks in advance!

Answer 1

Yes, the answer is $0$, but not really via your explanation.

You could put it in $0$-infinity indeterminate form in which you could note that $\cos{2t}$ is bounded and you are left with $1/t^{2}$ which goes to $0$ as x goes to infinity. (I let $t$ = $1/x$).

Answer 2

As Arturo says, we need the limits to exist to use the multiplication as you want to, to get around this we can do the following.

Define $u : = 1/x$ then we have $$\lim_{u\rightarrow \infty} \frac{1}{u^2}\cos(2u)$$

Now $\cos(2u)$ is bounded, while the $\frac{1}{u^2} \rightarrow 0$ and so the limit must be $0$.

Answer 3

You have that $$-1 \leq \cos \frac{2}{x} \leq 1$$

Then

$$-x^2f(x) \leq x^2f(x)\cos \frac{2}{x} \leq x^2f(x)$$

So you get for $x\to 0$

$$0 \leq \lim\limits_{x \to 0}x^2f(x)\cos \frac{2}{x} \leq 0$$