Answer
\[0\]
Work Step by Step
Let \[l=\lim_{x\rightarrow 0}\left(\frac{1-\cos x}{x^2+x}\right)\]
Which is $\frac{0}{0}$ form
Using L' Hopitals rule
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{(1-\cos x)'}{(x^2+x)'}\]
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{\sin x}{2x+1}\]
\[\Rightarrow l=\frac{\sin 0}{2(0)+1}\]
\[\Rightarrow l=\frac{0}{1}=0\]
Hence, \[\lim_{x\rightarrow 0}\left(\frac{1-\cos x}{x^2+x}\right)=0\]
