Limit value using serie definition

Question

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

Answer 1

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

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