Here is the question: How do I go about proving this? Thank you!
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Originally Posted by Phyxius117 Here is the question: How do I go about proving this? Thank you! Use the squeeze theorem and note that $\displaystyle \displaystyle -1 \le \sin \left(\frac{1}{x}\right) < 1 $ for all x.
Originally Posted by TheEmptySet Use the squeeze theorem and note that $\displaystyle \displaystyle -1 \le \sin \left(\frac{1}{x}\right) \le 1 $ for all x. ...for all $\displaystyle x \text{ such that }\displaystyle x \ne 0$. Luckily, the limit doesn't care about being at 0, just getting arbitrarily close to it.
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