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

Various exercises relating to orders of growth.

Use orders of growth to identify dominant terms and compute limits.
• Compute the limit: 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: 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: 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: 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:
$\sqrt {x^3+9}$ dominates $x$ $x$ dominates $\sqrt {x^3+9}$
• Compute the limit: (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:
• Compute the limit
$x$ dominantes $1$ as $x \rightarrow 0$ $1$ dominates $x$ as $x \rightarrow 0$
• Compute the limit:
$x^{0.01}$ dominates $(\ln x)^{100}$ as $x \rightarrow \infty$ $(\ln x)^{100}$ dominates $x^{0.01}$ as $x \rightarrow \infty$
• Compute the limit:
$x^{-0.01}$ dominates $(\ln x)^{100}$ as $x \rightarrow 0^+$ $(\ln x)^{100}$ dominates $x^{-0.01}$ as $x \rightarrow 0^+$
• Compute the limit: (Use the laws of exponents to write $x^2 = e^{2 \ln x}$.)

### Sample Quiz Questions

Arrange the functions 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.

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

Arrange the functions 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.

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

Arrange the functions 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.

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

Arrange the functions 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.

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