If I do this:

$\displaystyle \lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\frac{2\alpha}{5}} = \lim_{\alpha\rightarrow 0} \frac{5sin\alpha}{2\alpha} = \lim_{\alpha\rightarrow 0} \frac{5}{2}\cdot \frac{sin\alpha}{\alpha}$

Am I allowed to do this?

$\displaystyle \frac{5}{2} \cdot \lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\alpha} = \frac{5}{2} \cdot 1 = \frac{5}{2}$

Or did I do something in the very first steps incorrectly?