Q. 72 - 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.

$$\begin{gathered} u\left(x\right)=\ln \left(x^2+1\right), \\ v\left(x\right)=x^2+1 \end{gathered}$$

Step-by-Step Solution

Answer

The function $v\left(x\right)$ dominates $u\left(x\right)$ so $\underset{x\to \infty }{\lim }\left(x^2+1\right)=\infty$.

Step 1. Given information.

Consider the given question,

$$\begin{gathered} u\left(x\right)=\ln \left(x^2+1\right), \\ v\left(x\right)=x^2+1 \end{gathered}$$

Step 2. Use the limits.

Using the limits,

$\underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{\ln \left(x^2+1\right)}{x^2+1}$

Using L' Hospitals rule,

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

Step 3. Again use the limits.

Using the limits,

$\underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^2+1}{\ln \left(x^2+1\right)}$

Using L' Hospitals rule,

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