Question

determine the following limit.
\lim_{w \to \infty} \frac{20w^2 + 2w + 9}{\sqrt{25w^4 + w^3}}

select the correct choice, and, if necessary, fill in the answer box to complete your choice.

\bigcirc a. \\ \lim_{w \to \infty} \frac{20w^2 + 2w + 9}{\sqrt{25w^4 + w^3}} = \square \\ (simplify your answer.)

\bigcirc b. \\ the limit does not exist and is neither \infty nor -\infty.

Explanation:

Step1: Divide numerator and denominator by \( w^2 \)

For the numerator: \( \frac{20w^2 + 2w + 9}{w^2} = 20 + \frac{2}{w} + \frac{9}{w^2} \)
For the denominator: First, \( \sqrt{25w^4 + w^3} = \sqrt{w^4(25 + \frac{1}{w})} = w^2\sqrt{25 + \frac{1}{w}} \) (since \( w \to \infty \), \( w^2>0 \)), then divide by \( w^2 \): \( \frac{w^2\sqrt{25 + \frac{1}{w}}}{w^2} = \sqrt{25 + \frac{1}{w}} \)

Step2: Take the limit as \( w \to \infty \)

As \( w \to \infty \), \( \frac{2}{w} \to 0 \), \( \frac{9}{w^2} \to 0 \), and \( \frac{1}{w} \to 0 \).
So the limit becomes \( \frac{20 + 0 + 0}{\sqrt{25 + 0}} = \frac{20}{5} = 4 \)

Answer:

A. \( \lim\limits_{w\to\infty} \frac{20w^2 + 2w + 9}{\sqrt{25w^4 + w^3}} = 4 \)