Question

$\underset{x\to 2}{\lim }\frac{x^2-x-2}{x^2-2x+\sin \left(x-2\right)}$

Answer 1

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^2-x-2}{x^2-2x+\sin \left(x-2\right)} \\ =\underset{x\to 2}{\lim }\frac{\left(x-2\right)\left(x+1\right)}{x\left(x-2\right)+\sin \left(x-2\right)} \\ \mathrm{Let}y=x-2 \\ x\to 2 \\ \therefore y\to 0 \\ =\underset{y\to 0}{\lim }\frac{y\left(y+3\right)}{\left(y+2\right)y+\sin y} \\ \text{Dividing the numerator and the denominator by y:} \\ =\underset{y\to 0}{\lim }\frac{\left(y+3\right)}{\left(y+2\right)+{\displaystyle \frac{\sin y}{y}}} \\ =\frac{3}{2+1} \\ =\frac{3}{3}=1 \end{gathered}$$