Q. 43 - Calculus

Question

Calculate each limit in Exercises 35–80. 

$\underset{x\to -4}{lim}\frac{x^2+8x+16}{{(x+4)}^2(x+1)}$

Step-by-Step Solution

Answer

The limit is $-\frac{1}{3}$

Step 1. Given Information:

Given expression is $\underset{x\to -4}{lim}\frac{x^2+8x+16}{{(x+4)}^2(x+1)}$

We want to find the limit of given expression.

Step 2. Calculation:

The limit is calculated as below:

$\underset{x\to -4}{lim}\frac{x^2+8x+16}{{(x+4)}^2(x+1)}$

Obtain factor of numerator of expression we get

 $$\begin{gathered} \underset{x\to -4}{lim}\frac{{(x+4)}^2}{{(x+4)}^2(x+1)} \\ =\underset{x\to -4}{lim}\frac{1}{(x+1)} \end{gathered}$$

Now put limit we get

$\underset{x\to -4}{lim}\frac{1}{(x+1)}=\frac{1}{(-4+1)}=-\frac{1}{3}$