Exercises: Orders of Growth

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\bigmath }[1]{$\displaystyle #1$} \newcommand {\choicebreak }[0]{} \newcommand {\type }[0]{} \newcommand {\notes }[0]{} \newcommand {\keywords }[0]{} \newcommand {\offline }[0]{} \newcommand {\comments }[0]{\begin {feedback}} \newcommand {\multiplechoice }[0]{\begin {multipleChoice}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Various exercises relating to orders of growth.

Use orders of growth to identify dominant terms and compute limits.

Compute the limit: $\lim _{x \rightarrow \infty } \frac {e^x + x^2}{3 e^x + x^3} = \answer {\frac {1}{3}}$ The dominant terms are
$e^x$ in the numerator and $3 e^x$ in the denominator $e^x$ in the numerator and $x^3$ in the denominator $x^2$ in the numerator and $3 e^x$ in the denominator $x^2$ in the numerator and $x^3$ in the denominator
Compute the limit: $\lim _{x \rightarrow 0} \frac {e^x + x^2}{3 e^x + x^3} = \answer {\frac {1}{3}}$ The dominant terms are
$e^x$ in the numerator and $3 e^x$ in the denominator $e^x$ in the numerator and $x^3$ in the denominator $x^2$ in the numerator and $3 e^x$ in the denominator $x^2$ in the numerator and $x^3$ in the denominator
Compute the limit: $\lim _{x \rightarrow \infty } \frac {x + x^2}{3 x^4 + x^3} = \answer {0}$ The dominant terms are
$x$ in the numerator and $3 x^4$ in the denominator $x$ in the numerator and $x^3$ in the denominator $x^2$ in the numerator and $3 x^4$ in the denominator $x^2$ in the numerator and $x^3$ in the denominator
Compute the limit: $\lim _{x \rightarrow 0} \frac {x + x^2}{3 x^4 + x^3} = \answer {\infty }$ The dominant terms are
$x$ in the numerator and $3 x^4$ in the denominator $x$ in the numerator and $x^3$ in the denominator $x^2$ in the numerator and $3 x^4$ in the denominator $x^2$ in the numerator and $x^3$ in the denominator
Compute the limit: $\lim _{x \rightarrow \infty } \frac {\sqrt {x^3 + 9} - x}{\sqrt {x^3 + 9} + x} = \answer {1}$
$\sqrt {x^3+9}$ dominates $x$ $x$ dominates $\sqrt {x^3+9}$
Compute the limit: $\lim _{x \rightarrow \infty } \frac {\sqrt {x^3 + 9x^2} - x}{\sqrt {x^3 - 9x^2} + x} = \answer {1}$ (First identify whether the square-root term is dominant or not; when it is dominant, identify the dominant term inside the square root and neglect the non-dominant term.)
Compute the limit: $\lim _{x \rightarrow \infty } \frac { 2 x^3 + \sin x^5}{ 3 x^4 + \sin x^6} = \answer {0}.$
Compute the limit $\lim _{x \rightarrow 0} \frac {\sqrt {x + 1}}{\sqrt {x}} = \answer {\infty }.$
$x$ dominantes $1$ as $x \rightarrow 0$ $1$ dominates $x$ as $x \rightarrow 0$
Compute the limit: $\lim _{x \rightarrow \infty } \frac { (\ln x)^{100}}{x^{0.01}} = \answer {0}.$
$x^{0.01}$ dominates $(\ln x)^{100}$ as $x \rightarrow \infty$ $(\ln x)^{100}$ dominates $x^{0.01}$ as $x \rightarrow \infty$
Compute the limit: $\lim _{x \rightarrow 0^+} x^{0.01} (\ln x)^{100} = \answer {\infty }$
$x^{-0.01}$ dominates $(\ln x)^{100}$ as $x \rightarrow 0^+$ $(\ln x)^{100}$ dominates $x^{-0.01}$ as $x \rightarrow 0^+$
Compute the limit: $\lim _{x \rightarrow \infty } \frac {e^{\ln \ln x}+1}{x^2} = \answer {0}$ (Use the laws of exponents to write $x^2 = e^{2 \ln x}$ .)

Sample Quiz Questions

Arrange the functions $x^{x} \qquad \frac {e^{x}}{\ln {x}} \qquad \ln {x}$ in order from least rate of growth to greatest rate of growth as $x \rightarrow \infty$ . Compare on the basis of magnitude rather than sign, i.e., if a function is negative, take its absolute value first.

$\displaystyle x^{x} < \! \! < \frac {e^{x}}{\ln {x}} < \! \! < \ln {x}$ $\displaystyle \ln {x} < \! \! < x^{x} < \! \! < \frac {e^{x}}{\ln {x}}$ $\displaystyle \frac {e^{x}}{\ln {x}} < \! \! < \ln {x} < \! \! < x^{x}$ $\displaystyle x^{x} < \! \! < \ln {x} < \! \! < \frac {e^{x}}{\ln {x}}$ $\displaystyle \frac {e^{x}}{\ln {x}} < \! \! < x^{x} < \! \! < \ln {x}$ $\displaystyle \ln {x} < \! \! < \frac {e^{x}}{\ln {x}} < \! \! < x^{x}$

General Remarks:

Higher powers of $x$ grow faster at infinity than lower powers of $x$ .
As $x \rightarrow \infty$ , $\ln x$ goes to infinity slower than $x^p$ for any (presumably small) positive constant $p$ .
As $x \rightarrow \infty$ , $e^x$ goes to infinity faster than $x^n$ for any (presumably large) positive constant $n$ .
As $x \rightarrow \infty$ , $x^x$ goes to infinity faster than any exponential of the form $e^{cx}$ for any constant $c$ .

Arrange the functions $\frac {e^{- x}}{\ln {x}} \qquad x^{3} \ln {x} \qquad e^{x}$ in order from least rate of growth to greatest rate of growth as $x \rightarrow \infty$ . Compare on the basis of magnitude rather than sign, i.e., if a function is negative, take its absolute value first.

$\displaystyle \frac {e^{- x}}{\ln {x}} < \! \! < x^{3} \ln {x} < \! \! < e^{x}$ $\displaystyle e^{x} < \! \! < \frac {e^{- x}}{\ln {x}} < \! \! < x^{3} \ln {x}$ $\displaystyle x^{3} \ln {x} < \! \! < e^{x} < \! \! < \frac {e^{- x}}{\ln {x}}$ $\displaystyle \frac {e^{- x}}{\ln {x}} < \! \! < e^{x} < \! \! < x^{3} \ln {x}$ $\displaystyle x^{3} \ln {x} < \! \! < \frac {e^{- x}}{\ln {x}} < \! \! < e^{x}$ $\displaystyle e^{x} < \! \! < x^{3} \ln {x} < \! \! < \frac {e^{- x}}{\ln {x}}$

General Remarks:

Higher powers of $x$ grow faster at infinity than lower powers of $x$ .
As $x \rightarrow \infty$ , $\ln x$ goes to infinity slower than $x^p$ for any (presumably small) positive constant $p$ .
As $x \rightarrow \infty$ , $e^x$ goes to infinity faster than $x^n$ for any (presumably large) positive constant $n$ .

Arrange the functions $\left ( \ln \frac {1}{x} \right )^2 \qquad \frac {e^{- x}}{x^{3}} \ln {x} \qquad x^{3} e^{x}$ in order from least rate of growth to greatest rate of growth as $x \rightarrow 0^+$ . Compare on the basis of magnitude rather than sign, i.e., if a function is negative, take its absolute value first.

$\displaystyle \left ( \ln \frac {1}{x} \right )^2 < \! \! < \frac {e^{- x}}{x^{3}} \ln {x} < \! \! < x^{3} e^{x}$ $\displaystyle x^{3} e^{x} < \! \! < \left ( \ln \frac {1}{x} \right )^2 < \! \! < \frac {e^{- x}}{x^{3}} \ln {x}$ $\displaystyle \frac {e^{- x}}{x^{3}} \ln {x} < \! \! < x^{3} e^{x} < \! \! < \left ( \ln \frac {1}{x} \right )^2$ $\displaystyle \left ( \ln \frac {1}{x} \right )^2 < \! \! < x^{3} e^{x} < \! \! < \frac {e^{- x}}{x^{3}} \ln {x}$ $\displaystyle \frac {e^{- x}}{x^{3}} \ln {x} < \! \! < \left ( \ln \frac {1}{x} \right )^2 < \! \! < x^{3} e^{x}$ $\displaystyle x^{3} e^{x} < \! \! < \frac {e^{- x}}{x^{3}} \ln {x} < \! \! < \left ( \ln \frac {1}{x} \right )^2$

General Remarks:

Lower powers of $x$ grow faster as $x \rightarrow 0^+$ than higher powers of $x$ .
As $x \rightarrow 0^+$ , $-\ln x = \ln x^{-1}$ goes to $\infty$ slower than $x^{-p}$ for any (presumably small) positive $p$ .
As $x \rightarrow 0^+$ , $e^{x} \rightarrow 1$ and so does not influence the growth rate.
As $x \rightarrow 0^+$ , $e^{-x} \rightarrow 1$ and so does not influence the growth rate.

Arrange the functions $\frac {x^{3} e^{x}}{\ln {x}} \qquad \frac {e^{- x}}{x^{3}} \qquad \frac {e^{- x}}{x^{3} \ln {x}}$ in order from least rate of growth to greatest rate of growth as $x \rightarrow 0^+$ . Compare on the basis of magnitude rather than sign, i.e., if a function is negative, take its absolute value first.

$\displaystyle \frac {x^{3} e^{x}}{\ln {x}} < \! \! < \frac {e^{- x}}{x^{3}} < \! \! < \frac {e^{- x}}{x^{3} \ln {x}}$ $\displaystyle \frac {e^{- x}}{x^{3} \ln {x}} < \! \! < \frac {x^{3} e^{x}}{\ln {x}} < \! \! < \frac {e^{- x}}{x^{3}}$ $\displaystyle \frac {e^{- x}}{x^{3}} < \! \! < \frac {e^{- x}}{x^{3} \ln {x}} < \! \! < \frac {x^{3} e^{x}}{\ln {x}}$ $\displaystyle \frac {x^{3} e^{x}}{\ln {x}} < \! \! < \frac {e^{- x}}{x^{3} \ln {x}} < \! \! < \frac {e^{- x}}{x^{3}}$ $\displaystyle \frac {e^{- x}}{x^{3}} < \! \! < \frac {x^{3} e^{x}}{\ln {x}} < \! \! < \frac {e^{- x}}{x^{3} \ln {x}}$ $\displaystyle \frac {e^{- x}}{x^{3} \ln {x}} < \! \! < \frac {e^{- x}}{x^{3}} < \! \! < \frac {x^{3} e^{x}}{\ln {x}}$

General Remarks:

Lower powers of $x$ grow faster as $x \rightarrow 0^+$ than higher powers of $x$ .
As $x \rightarrow 0^+$ , $-\ln x = \ln x^{-1}$ goes to $\infty$ slower than $x^{-p}$ for any (presumably small) positive $p$ .
As $x \rightarrow 0^+$ , $e^{x} \rightarrow 1$ and so does not influence the growth rate.
As $x \rightarrow 0^+$ , $e^{-x} \rightarrow 1$ and so does not influence the growth rate.