## Calculus: Early Transcendentals (2nd Edition)

The increasing order is $$\ln x,\quad x^3,\quad 2^x,\quad x^x.$$
The increasing order is $$\ln x,\quad x^3,\quad 2^x,\quad x^x.$$ To prove this we calculate: 1) $$\lim_{x\to\infty}\frac{\ln x}{x^3}=\left[\frac{\infty}{\infty}\right][\text{Apply L'H Rule}]=\lim_{x\to\infty}\frac{(\ln x)'}{(x^3)'}=\lim_{x\to\infty}\frac{\frac{1}{x}}{3x^2}=\lim_{x\to\infty}\frac{1}{3x^3}=\left[\frac{1}{\infty}\right]=0.$$ so $\ln x$ grows slower than $x^3$. 2) $$\lim_{x\to\infty}\frac{x^3}{2^x}=\left[\frac{\infty}{\infty}\right][\text{Apply L'H Rule}]=\lim_{x\to\infty}\frac{(x^3)'}{(2^x)'}=\lim_{x\to\infty}\frac{3x^2}{2^x\ln 2}=\left[\frac{\infty}{\infty}\right][\text{Apply L'H Rule}]=\lim_{x\to\infty}\frac{(3x^2)'}{(2^x\ln 2)'}=\lim_{x\to \infty}\frac{6x}{2^x\ln^22}=\left[\frac{\infty}{\infty}\right][\text{Apply L'H Rule}]=\lim_{x\to\infty}\frac{(6x)'}{(2^x\ln^2 2)'}=\lim_{x\to \infty}\frac{6}{2^x\ln^32}=\left[\frac{6}{\infty}\right]=0,$$ so $x^3$ grows slower than $2^x$. 3) $$\lim_{x\to\infty}\frac{2^x}{x^x}=\lim_{x\to\infty}\left(\frac{2}{x}\right)^x=\left[\left(\frac{2}{\infty}\right)^\infty\right]=\left[0^\infty\right]=0,$$ which means that $2^x$ grows slower than $x^x$