Question

Evaluate the limit if it exists.

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

Short answer

The value of limit of the given function is$\frac{3}{5}.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

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

Step 2. Formula used.

The given function is a rational number. We can find the limit by cancelling a common factor.

Step 3. Simplification.

Given,

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

$\begin{array}{c}\underset{x\to -4}{\lim }\frac{x^2+5x+4}{x^2+3x-4}=\frac{x^2+(4+1)x+4}{x^2+(4-1)x-4}\\ =\frac{x^2+4x+x+4}{x^2+4x-x-4}\\ =\frac{x(x+4)+1(x+4)}{x(x+4)-1(x+4)}\\ =\frac{(x+4)(x+1)}{(x+4)(x-1)}\\ =\frac{x+1}{x-1}\\ \end{array}$

Now, applying the value:

$\begin{array}{c}=\frac{-4+1}{-4-1}\\ =\frac{-3}{-5}\\ =\frac{3}{5}\end{array}$

Therefore, our final answer is$\frac{3}{5}.$