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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!

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No, you should use the squeeze law. – M.B. Mar 25 '12 at 23:29
No, that's an invalid use of the limit laws. The multiplication law says: if $\lim\limits_{x\to a}f(x) =a$ and $\lim\limits_{x\to a}g(x) = b$, then $\lim\limits_{x\to a}f(x)g(x) = ab$. In order to be able to say the limit of the product is equal to the product of the limits, you need both limits to exist. Does $\lim\limits_{x\to 0}\cos(2/x)$ exist? If the answer is "no", then what you are trying to do is invalid. – Arturo Magidin Mar 25 '12 at 23:33

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


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

So you get for $x\to 0$

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

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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 $t$ = $1/x$).

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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 $0$.

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Thanks very much! – JackReacher Mar 26 '12 at 0:59

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