Q. 74 - Calculus

Question

A function f dominates another function g as $x\to \infty$ if $f\left(x\right),g\left(x\right)$ both grow without bound as $x\to \infty$ and if $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$.

Intuitively, f dominates g as $x\to \infty$ if $f\left(x\right)$ is very much larger than $g\left(x\right)$for very large values of x. Use limits to determine whether $u\left(x\right)$ dominates $v\left(x\right)$ or $v\left(x\right)$ dominates $u\left(x\right)$ or neither.

$$\begin{gathered} u\left(x\right)=0.001x^2-100x, \\ v\left(x\right)=100{\log }_{3}x \end{gathered}$$

Step-by-Step Solution

Answer

The function $u\left(x\right)$ dominates $\mathrm{v}\left(\mathrm{x}\right)$ so $\underset{x\to \infty }{\lim }\frac{x}{{\log }_3 x}=-\infty$.

Step 1. Given information.

Consider the given question,

$$\begin{gathered} u\left(x\right)=0.001x^2-100x, \\ v\left(x\right)=100{\log }_{3}x \end{gathered}$$

Step 2. Use the limits.

Using the limits,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{0.001e^2-100x}{100{\log }_{3}x} \\ \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\left(\frac{0.00001e^2}{{\log }_{3}x}-\frac{x}{{\log }_{3}x}\right) \\ \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x}{{\log }_{3}x} \\ \underset{x\to \infty }{\lim }\frac{x}{{\log }_{3}x}=0-\infty \\ \underset{x\to \infty }{\lim }\frac{x}{{\log }_{3}x}=-\infty \end{gathered}$$

Step 3. Again use the limits.

Using the limits,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{100{\log }_{3}x}{0.001e^2-100x} \\ \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_{3}x}{0.00001e^2-x} \\ \underset{x\to \infty }{\lim }\frac{{\log }_{3}x}{0.00001e^2-x}=0 \end{gathered}$$