Question

Estimating limits numerically and graphically.

Use a table of values to estimate the limit. Then use a graphing device to confirm your result graphically.

$\underset{x\to -\infty }{\lim }\frac{\sqrt{x^2+4x}}{4x+1}$

Short answer

The value of limit of the given function is$\frac{1}{4}.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

$\underset{x\to -\infty }{\lim }\frac{\sqrt{x^2+4x}}{4x+1}$

Step 2. Formula used.

Here, we simplify the fractions.

Step 3. Simplification.

Given,

$\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{\sqrt{x^2+4x}}{4x+1}\\ =\underset{x\to -\infty }{\lim }\frac{\sqrt{x^2(1+\frac{4x}{x^2})}}{4x+1}\\ =\underset{x\to -\infty }{\lim }\frac{x\sqrt{1+\frac{4}{x}}}{4x+1}\\ =\underset{x\to -\infty }{\lim }\frac{x\sqrt{1+\frac{4}{x}}}{x\left(4+\frac{1}{x}\right)}\\ =\underset{x\to -\infty }{\lim }\frac{\sqrt{1+\frac{4}{x}}}{4+\frac{1}{x}}\end{array}$

Now, we put the limit:

$\begin{array}{c}=\frac{\sqrt{1+\frac{4}{\infty }}}{4+\frac{1}{\infty }}\\ =\frac{\sqrt{1+0}}{4+0}\\ =\frac{\sqrt{1}}{4}\\ =\frac{1}{4}\end{array}$

Therefore, our final answer is $\frac{1}{4}.$