Chapter 3: Q. 72 (page 311)

A function f dominates another function g as $x \to \infty$ if $f (x), g (x)$ both grow without bound as $x \to \infty$ and if $\lim_{x \to \infty} \frac{f (x)}{g (x)} = \infty$ .
Intuitively, f dominates gas $x \to \infty$ if $f (x)$ is very much larger than $g (x)$ for very large values of x. Use limits to determine whether $u (x)$ dominates $v (x)$ or $v (x)$ dominates $u (x)$ or neither.
$u (x) = \ln (x^{2} + 1), v (x) = x^{2} + 1$

Short Answer

Expert verified

The function $v (x)$ dominates $u (x)$ so localid="1648749563681" $\lim_{x \to \infty} (x^{2} + 1) = \infty$ .

Step by step solution

Step 1. Given information.

Consider the given question,

$u (x) = \ln (x^{2} + 1), v (x) = x^{2} + 1$

Step 2. Use the limits.

Using the limits,

$\lim_{x \to \infty} \frac{u (x)}{v (x)} = \lim_{x \to \infty} \frac{\ln (x^{2} + 1)}{x^{2} + 1}$

Using L' Hospitals rule,

$\lim_{x \to \infty} \frac{u (x)}{v (x)} = \lim_{x \to \infty} \frac{\frac{1}{x^{2} + 1} \cdot 2 x}{2 x} \lim_{x \to \infty} \frac{u (x)}{v (x)} = \lim_{x \to \infty} (\frac{1}{x^{2} + 1} . \frac{2 x}{2 x}) \lim_{x \to \infty} \frac{u (x)}{v (x)} = \lim_{x \to \infty} (\frac{1}{x^{2} + 1}) \lim_{x \to \infty} (\frac{1}{x^{2} + 1}) = \frac{1}{\infty} \lim_{x \to \infty} (\frac{1}{x^{2} + 1}) = 0$

Step 3. Again use the limits.

Using the limits,

$\lim_{x \to \infty} \frac{v (x)}{u (x)} = \lim_{x \to \infty} \frac{x^{2} + 1}{\ln (x^{2} + 1)}$

Using L' Hospitals rule,

localid="1648749543558" $\lim_{x \to \infty} \frac{v (x)}{u (x)} = \lim_{x \to \infty} \frac{2 x}{\frac{1}{x^{2} + 1} \cdot 2 x} \lim_{x \to \infty} \frac{v (x)}{u (x)} = \lim_{x \to \infty} (x^{2} + 1) \cdot \frac{2 x}{2 x} \lim_{x \to \infty} \frac{v (x)}{u (x)} = \lim_{x \to \infty} (x^{2} + 1) \lim_{x \to \infty} (x^{2} + 1) = \infty$