Q. 46 - Calculus

Question

Calculate each limit in Exercises 35–80. 

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

Step-by-Step Solution

Answer

The limit is 0.

Step 1. Given Information:

Given expression: $\underset{x\to -4}{\lim }\frac{x+4}{x^2+8x+16}$

We want to find the limit of given expression.

Step 2. Solution:

Put limit we get: $\underset{x\to -4}{\lim }\frac{x+4}{x^2+8x+16}=\frac{-4+4}{{(-4)}^2+8(-4)+16}=\frac{0}{0}$

Using L'Hospital rule we get:

$$\begin{gathered} \underset{x\to -4}{\lim }\frac{{\displaystyle \frac{d}{dx}}(x+4)}{{\displaystyle \frac{d}{dx}}(x^2+8x+16)} \\ =\underset{x\to -4}{\lim }\frac{1}{2x+8} \end{gathered}$$

Now put limit we get

$\underset{x\to -4}{\lim }\frac{1}{2x+8}=\frac{1}{2(-4)+8}=\frac{1}{0}$

Again Using L'Hospital rule we get:

$\underset{x\to -4}{\lim }\frac{1}{2x+8}=\underset{x\to -4}{\lim }\frac{{\displaystyle \frac{d}{dx}}1}{{\displaystyle \frac{d}{dx}}(2x+8)}=\underset{x\to -4}{\lim }\frac{0}{2}=\underset{x\to -4}{\lim }0=0$

Hence the limit is 0.