Answer
$$\lim _{x \rightarrow 0} \frac{x^{2}-x+\sin x}{2 x}=0$$
Work Step by Step
Given $$\lim _{x \rightarrow 0} \frac{x^{2}-x+\sin x}{2 x}$$
So, we get
\begin{aligned}L&=\lim _{x \rightarrow 0} \frac{x^{2}-x+\sin x}{2 x}\\
&=\lim _{x \rightarrow 0}\left(\frac{x^{2}}{2 x}-\frac{x}{2 x}+\frac{\sin x}{2 x}\right)\\
&=\lim _{x \rightarrow 0}\left(\frac{1}{2} \cdot x-\frac{1}{2}+\frac{1}{2} \cdot \frac{\sin x}{x}\right)\\
&=
\lim _{x \rightarrow 0} \frac{1}{2} \cdot x-\lim _{x \rightarrow 0} \frac{1}{2}+\lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{\sin x}{x}\\
&=0- \frac{1}{2}+\frac{1}{2} \cdot \lim _{x \rightarrow 0} \frac{\sin x}{x}\\
&=-\frac{1}{2}+\frac{1}{2} \cdot 1\\
&=-\frac{1}{2}+\frac{1}{2}\\
&=0
\end{aligned}
