Why is the limit 1?
You could use used L'hospital's rule to solve the limit.
$\displaystyle \lim \limits_{x \rightarrow 0} \dfrac{f(x)}{g(x)} = \lim \limits_{x \rightarrow 0} \dfrac{f'(x)}{g'(x)}$
$\displaystyle \lim \limits_{\sigma x \rightarrow 0} \dfrac{\sin{\frac{\sigma x}{2}}}{\frac{\sigma x}{2}}$ $\displaystyle = \lim \limits_{\sigma x \rightarrow 0} \dfrac{0.5 \cos \frac{\sigma x}{2}}{0.5}$
Substitute in $\displaystyle \sigma x = 0$ and you will get your answer.
Whether that is valid or not depends upon exactly how you have proved that the derivative of sin(x) is cos(x). Most textbooks use the limit $\displaystyle \displaytype\lim_{x\to 0}\frac{sin(\theta)}{\theta}= 1$. If you already have that then the given limit follows immediately by letting $\displaystyle \theta= \frac{\delta x}{2}$.