Question

In Problems 1– 6, find each limit

$\underset{x\to -1}{\lim }\frac{x^2-4x-5}{x^3+1}$

Short answer

The value is$-2$

Step-by-step solution

Step 1. Rewriting the limit

We can write the given limit in simplified form as,

$\underset{x\to -1}{\lim }\frac{x^2-4x-5}{x^3+1}=\underset{x\to -1}{\lim }\frac{(x+1)(x-5)}{(x+1)\left(x^2-x+1\right)}$

Cancelling common terms, the expression becomes,

$$\begin{gathered} =\underset{x\to -1}{\lim }\frac{x-5}{x^2-x+1} \\ =\frac{{\lim }_{x\to -1}(x-5)}{{\lim }_{x\to -1}\left(x^2-x+1\right)} \end{gathered}$$

Step 2. Value of the limit

On simplifying,

$$\begin{gathered} \frac{{\lim }_{x\to -1}}{{\lim }_{x\to -1}\left(x^2-x+1\right)}=\frac{{\lim }_{x\to -1}x-{\lim }_{x\to -1}5}{{\lim }_{x\to -1}{x}^2-{\lim }_{x\to -1}x+{\lim }_{x\to -1}1} \\ =\frac{-1-{\lim }_{x\to -1}5}{(-1{)}^2-(-1)+{\lim }_{x\to -1}1} \\ =\frac{-1-5}{(-1{)}^2-(-1)+1} \\ =\frac{-6}{3} \\ =-2 \end{gathered}$$