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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.2012-06-17

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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?2012-06-17
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    @mystycs What do you mean?2012-06-17
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    Oh nevermind if its cos it has to be between -1 and 1 i got it2012-06-17
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    So you just plugin between two values and just prove it correct?2012-06-17
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    To prove that $-1 \le \cos(1/x^2) \le 1$? It follows from the identity $sin^2 + cos^2 = 1$.2012-06-17
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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.2012-06-17
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    @ArturoMagidin Right. Thanks.2012-06-17