Answer
$\infty $
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} \cr
& {\text{Evaluate the limit by direct substitution}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} = \frac{{{{\left( 1 \right)}^2} + 2\left( 1 \right)}}{{{{\left( 1 \right)}^2} - 2\left( 1 \right) + 1}} = \frac{3}{0}{\text{ }} \cr
& {\text{Finding the one sided limits }} \cr
& \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}}{\text{ and }}\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} \cr
& \cr
& *\mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}},{\text{ taking }}x = 0.999 \cr
& \approx \frac{{{{\left( {0.999} \right)}^2} + 2\left( {0.999} \right)}}{{{{\left( {0.999} \right)}^2} - 2\left( {0.999} \right) + 1}} \approx 2996001,{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} = + \infty \cr
& \cr
& *\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}},{\text{ taking }}x = 1.001 \cr
& \approx \frac{{{{\left( {1.001} \right)}^2} + 2\left( {1.001} \right)}}{{{{\left( {1.001} \right)}^2} - 2\left( {1.001} \right) + 1}} \approx 3004001,{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} = + \infty \cr
& \cr
& \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} = \infty ,{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 2x}}{{{x^2} - 2x + 1}} = \infty \cr} $$
