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How can I argue that $$\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right) = 0$$

I understand I have to use a squeeze theorem and that one piece goes to zero but I'm not sure how to tackle this problem to show on a test.

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Squeeze theorem is a big hint. Since you know you have to apply squeeze theorem, that means you need to find upper and lower bound for your function, for which you are trying to find the limit. Now, look at your function and figure out whether you can give upper and lower bounds for it. – TenaliRaman Jun 17 '12 at 22:08
up vote 6 down vote accepted

Use $-1 \le \cos(\frac{1}{x^2}) \le 1$ and multiply through by $x^2$. Since $x^2 \ge 0$, the inequalities remain valid.

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I thought it was between 0? – soniccool Jun 17 '12 at 22:13
@mystycs What do you mean? – Cocopuffs Jun 17 '12 at 22:18
Oh nevermind if its cos it has to be between -1 and 1 i got it – soniccool Jun 17 '12 at 22:18
So you just plugin between two values and just prove it correct? – soniccool Jun 17 '12 at 22:20
To prove that $-1 \le \cos(1/x^2) \le 1$? It follows from the identity $sin^2 + cos^2 = 1$. – Cocopuffs Jun 17 '12 at 22:23

We have that $-1 \leq \cos (1/x^2) \leq 1$ for any $x$. So $-x^2 \leq x^2\cos(1/x^2) \leq x^2$. Therefore $$ \lim_{x \to 0} -x^2 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq \lim_{x \to 0} x^2. $$ But we have that $$ \lim_{x \to 0} x^2 = 0 $$ and $$ \lim_{x \to 0} -x^2 = 0. $$

So $$ 0 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq 0 $$ and therefore by the squeeze theorem, $$ \lim_{x \to 0} x^2\cos(1/x^2) = 0. $$

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Please use \cos rather than just cos inside equations. – Arturo Magidin Jun 17 '12 at 22:38
@ArturoMagidin Right. Thanks. – Eugene Jun 17 '12 at 22:39

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