1. ## Limit x-> 0

lim x->0 ( X2-x+sinx)/(2x)

I can't seem to figure this one out, thanks

2. Hello, Soulwraith56!

You're expected to know that: . $\lim_{\theta\to0} \frac{\sin\theta}{\theta} \:=\:1$

$\lim_{x\to0}\frac{x^2-x+\sin x}{2x}$

We have: . $\lim_{x\to0}\left(\frac{x^2}{2x} - \frac{x}{2x} + \frac{\sin x}{2x}\right)$

. . $=\;\;\lim_{x\to0}\left(\frac{1}{2}x - \frac{1}{2} + \frac{1}{2}\!\cdot\!\frac{\sin x}{x}\right) \;\;=\;\;\frac{1}{2}(0) - \frac{1}{2} + \frac{1}{2}(1) \;\;=\;\;0$