Q. 71 - 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.

$$\begin{gathered} u\left(x\right)={\log }_2\left(x\right), \\ v\left(x\right)={\log }_{30}\left(x\right) \end{gathered}$$

Step-by-Step Solution

Answer

Neither functions$u\left(x\right)$ or $v\left(x\right)$ dominates each other.

Step 1. Given information.

Consider the given question,

$$\begin{gathered} u\left(x\right)={\log }_2\left(x\right), \\ v\left(x\right)={\log }_{30}\left(x\right) \end{gathered}$$

Step 2. Use the limits.

Using the limits,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_{2}x}{{\log }_{30}x} \\ \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_{x}30}{{\log }_{x}2} \\ \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_x\left(2\times 15\right)}{{\log }_{x}2} \\ \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_{x}2+{\log }_{x}15}{{\log }_{x}2} \\ \underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\left(1+\frac{{\log }_{x}15}{{\log }_{x}2}\right) \\ \underset{x\to \infty }{\lim }\left(1+\frac{{\log }_{x}15}{{\log }_{x}2}\right)=1 \end{gathered}$$

Step 3. Again use the limits.

Using the limits,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_{30}x}{{\log }_{2}x} \\ \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_{2}x}{{\log }_{30}x} \\ \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_x\left({\displaystyle \frac{60}{30}}\right)}{{\log }_{x}30} \\ \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\log }_{x}60-{\log }_{x}30}{{\log }_{x}30} \\ \underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\left(\frac{{\log }_{x}60}{{\log }_{x}30}-1\right) \\ \underset{x\to \infty }{\lim }\left(\frac{{\log }_{x}60}{{\log }_{x}30}-1\right)=-1 \end{gathered}$$

Hence, neither functions dominates each other.