Q. 73 - 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 $v\left(x\right)$ so $\underset{x\to \infty }{\lim }\frac{0.001e^{0.001x}}{100x^{100}}=\infty$.

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,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{0.001e^{0.001x}}{100x^{100}} \\ \underset{x\to \infty }{\lim }\frac{0.001e^{0.001x}}{100x^{100}}=\infty \end{gathered}$$

Again, using the limits,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{100x^{100}}{0.001e^{0.001x}} \\ \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{10000e^{0.001x}}{x^{100}} \\ \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=10000\underset{x\to \infty }{\lim }\frac{e^{0.001x}}{x^{100}} \\ 10000\underset{x\to \infty }{\lim }\frac{e^{0.001x}}{x^{100}}=0 \end{gathered}$$