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