Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
No, you should use the squeeze law. –  M.B. Mar 25 '12 at 23:29
1  
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
add comment

3 Answers

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 $x\to 0$

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

share|improve this answer
add comment

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 $t$ = $1/x$).

share|improve this answer
add comment

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 $0$.

share|improve this answer
    
Thanks very much! –  JackReacher Mar 26 '12 at 0:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.