Q. 68 - 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.

$$\begin{gathered} u\left(x\right)=x^2, \\ v\left(x\right)=2^x \end{gathered}$$

Step-by-Step Solution

Answer

The function $v\left(x\right)$ dominates $u\left(x\right)$ so $\underset{x\to \infty }{\lim }\frac{2^x}{x^2}=\infty$.

Step 1. Given information.

Consider the given question,

$$\begin{gathered} u\left(x\right)=x^2, \\ v\left(x\right)=2^x \end{gathered}$$

Step 2. Use the limits.

Using the limits,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^2}{2^x} \\ =\underset{x\to \infty }{\lim }\left(\frac{1}{{\displaystyle \frac{2^x}{x^2}}}\right) \\ =\frac{1}{\infty } \\ =0 \end{gathered}$$

Again, $$\begin{gathered} \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{2^x}{x^2} \\ \underset{x\to \infty }{\lim }\frac{2^x}{x^2}=\infty \end{gathered}$$