Question

$\underset{x\to -1}{\lim }\frac{x^2-x-2}{\left(x^2+x\right)+\sin \left(x+1\right)}$

Answer 1

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