Para provar que o limite existe, realizaremos os cálculos abaixo:

\(\begin{align} & 1-\delta <x<1+\delta \\ & |x-1|<\delta \\ & 2-\frac{1}{3}<{{x}^{2}}+x<2+\frac{1}{3} \\ & |{{x}^{2}}+x-2|<\frac{1}{3} \\ & |(x-1)(x+2)|<\frac{1}{3} \\ & |x-1||x+2|<\frac{1}{3} \\ & \\ & |x-1|<\delta \\ & |x+2|=|x-1+3|\le |x-1|+|3|<\delta +3 \\ & {{\delta }^{2}}+3\delta <\frac{1}{3} \\ & 4\delta <\frac{1}{3} \\ & \delta <\frac{1}{12} \\ & \delta =0,88 \\ & \\ & \underset{x\to 1}{\mathop{\lim }}\,{{x}^{2}}+x=2 \\ \end{align}\ \)