Chapter 9.3, Problem 7E

To determine

To show that $\ln x$ grows slower than the given function.

Explanation of Solution

Given information:

The given function is $\sqrt[3]{x}$ .

Formula used:

g grows than f as $x\to \infty \text{ if }\underset{x\to \infty }{\lim }\frac{g\left(x\right)}{f\left(x\right)}=0$

Proof:

Here,

  $f\left(x\right)=\sqrt[3]{x},g\left(x\right)=\ln x$

Now substitute $f\left(x\right)$ and $g\left(x\right)$

  $\underset{x\to \infty }{\lim }\frac{\ln x}{\sqrt[3]{x}}$

The limit is $\frac{\infty }{\infty }$ type, so use l Hopital’s Rule and take derivative of numerator and of denominator.

  $\begin{array}{l}or,\underset{x\to \infty }{\lim }\frac{\ln x}{\sqrt[3]{x}}\\ =\underset{x\to \infty }{\lim }\frac{\frac{1}{x}}{\frac{1}{3x^{\frac{2}{3}}}}\\ =\underset{x\to \infty }{\lim }\frac{3x^{\frac{2}{3}}}{x}\left[\text{multiply numerator and denominator by}3x^{\frac{2}{3}}\right]\\ =\underset{x\to \infty }{\lim }\frac{3}{x^{\frac{1}{3}}}\\ =0\end{array}$

Hence,

  $\ln x$ Grows slower than $\sqrt[3]{x}$ .