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I need to find the limit value of $$\lim_{x\to\:0}\frac{sin(\frac{1}x)}{\frac{1}x}$$

I wanted to do it with the serie definition of sinus and I come to the result:

$$ 1 -\lim_{x\to\:0} \sum_{i=0}^{\infty} (-1)^n{\frac{1}{x^{2n}}}$$

The limit value by substituion of:

$$ \frac{1}{x}=y$$ in the original limit is so $0$, this means that the limit value of the serie should be $1$, however I cannot find a method to prove that. Can someone help me? Thanks

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  • $\begingroup$ Welcome to stackexchange. The series definition is not a good strategy. After the substitution, the limit as $ y \to \infty$ is $0$, as you say. That's a proof, as long as you know why that limit is $0$. $\endgroup$ – Ethan Bolker Feb 3 '18 at 20:36
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png $\endgroup$ – user Feb 5 '18 at 23:38
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Note that $\sin\left( \frac1x \right) $ has not a series expansion for $x\to 0$.

You can show the limit easily by squeeze theorem, indeed

$$-x\le x \sin\left( \frac1x \right)\le x$$

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