Q. 70 - 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 data-custom-editor="chemistry" $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$.

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 data-custom-editor="chemistry" $\mathrm{v}\left(\mathrm{x}\right)$ or $v\left(x\right)$ dominates $u\left(x\right)$ or neither.

$$\begin{gathered} u\left(x\right)=x^{10}+3, \\ v\left(x\right)={10}^x+3 \end{gathered}$$

Step-by-Step Solution

Answer

The function $u\left(x\right)$ dominates $v\left(x\right)$ so $\underset{x\to \infty }{\lim }\frac{x^{10}+3}{{10}^x+3}=\infty$.

Step 1. Given information.

Consider the given question,

$$\begin{gathered} u\left(x\right)=x^{10}+3, \\ v\left(x\right)={10}^x+3 \end{gathered}$$

Step 2. Use the limits.

Using the limits,

$\underset{x\to \infty }{\lim }\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^{10}+3}{{10}^x+3}$

As $x^{10}>{10}^x$,

$\underset{x\to \infty }{\lim }\frac{x^{10}+3}{{10}^x+3}=\infty$

Again, $\underset{x\to \infty }{\lim }\frac{v\left(x\right)}{u\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{10}^x+3}{x^{10}+3}$

As $x^{10}>{10}^x$,

$\underset{x\to \infty }{\lim }\frac{{10}^x+3}{x^{10}+3}=0$