MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 36

Limits

When a limit looks like it is coming out to $\frac{0}{0}$ it could really be coming out to anything.

$\underset{x\to 0}{\text{lim}}\frac{5x}{x}=\frac{0}{0}\text{??}$ BAD $$\underset{x\to 0}{\text{lim}}\frac{5x}{x}=\underset{x\to 0}{\text{lim}}5=5\text{.}$$

$\underset{x\to 0}{\text{lim}}\frac{642x}{x}=\frac{0}{0}\text{??}$ BAD $$\underset{x\to 0}{\text{lim}}\frac{642x}{x}=\underset{x\to 0}{\text{lim}}642=642\text{.}$$

$\underset{x\to 0}{\text{lim}}\frac{x^2}{x}=\frac{0}{0}\text{??}$ BAD $$\underset{x\to 0}{\text{lim}}\frac{x^2}{x}=\underset{x\to 0}{\text{lim}}x=0\text{.}$$

$\underset{x\to 0}{\text{lim}}\frac{x}{x^2}=\frac{0}{0}\text{??}$ BAD $$\underset{x\to 0}{\text{lim}}\frac{x}{x^2}=\underset{x\to 0}{\text{lim}}\frac{1}{x}=\text{UNDEFINED}$$ since $$\begin{array}{ccc}{\displaystyle \underset{x\to 0^{+}}{\text{lim}}\frac{x}{x^2}}& =& {\displaystyle \underset{x\to 0^{+}}{\text{lim}}\frac{1}{x}=\infty ,}\\ {\displaystyle \underset{x\to 0^{-}}{\text{lim}}\frac{x}{x^2}}& =& {\displaystyle \underset{x\to 0^{-}}{\text{lim}}\frac{1}{x}=-\infty \text{.}}\end{array}$$

Bad limits $\frac{0}{0}$ and $\frac{\infty }{\infty }$ and L'Hopital's rule

If $$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}\text{???}\text{BAD}$$ or $$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\infty }{\infty }\text{???}\text{BAD}$$ then sometimes it works to use $$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\text{lim}}\frac{f\prime \left(x\right)}{g\prime \left(x\right)}\text{.}$$

$\underset{x\to 1}{\text{lim}}\frac{\text{ln}\hspace{0.17em}x}{x-1}$

First: $$\underset{x\to 1}{\text{lim}}\frac{\text{ln}\hspace{0.17em}x}{x-1}=\frac{0}{0}\text{???}\text{BAD}$$ Try L'Hopital's rule: $$\underset{x\to 1}{\text{lim}}\frac{\text{ln}\hspace{0.17em}x}{x-1}=\underset{x\to 1}{\text{lim}}\frac{\frac{1}{x}}{1}=\underset{x\to 1}{\text{lim}}\frac{1}{x}=\frac{1}{1}=1\text{.}$$

$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\hspace{0.17em}x}{e^x}$

First: $$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\hspace{0.17em}x}{e^x}=\frac{\infty }{\infty }\text{???}\text{BAD}$$ Try L'Hopital's rule: $$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\hspace{0.17em}x}{e^x}=\underset{x\to \infty }{\text{lim}}\frac{\frac{1}{x}}{e^x}=\underset{x\to \infty }{\text{lim}}\frac{1}{x{e}^x}=0\text{.}$$

$\underset{x\to \infty }{\text{lim}}\frac{{\left(\text{ln}\hspace{0.17em}x\right)}^3}{x^2}$

First: $$\underset{x\to \infty }{\text{lim}}\frac{{\left(\text{ln}\hspace{0.17em}x\right)}^2}{x^2}=\frac{\infty }{\infty }\text{???}\text{BAD}$$ Try L'Hopital's rule: $$\underset{x\to \infty }{\text{lim}}\frac{{\left(\text{ln}\hspace{0.17em}x\right)}^2}{x^2}=\underset{x\to \infty }{\text{lim}}\frac{3{\left(\text{ln}\hspace{0.17em}x\right)}^2\frac{1}{x}}{2x}=\underset{x\to \infty }{\text{lim}}\frac{3{\left(\text{ln}\hspace{0.17em}x\right)}^2}{2x^2}=\frac{\infty }{\infty }\text{???}$$ Try again: $$\underset{x\to \infty }{\text{lim}}\frac{3{\left(\text{ln}\hspace{0.17em}x\right)}^2}{2x^2}=\underset{x\to \infty }{\text{lim}}\frac{6\left(\text{ln}\hspace{0.17em}x\right)\frac{1}{x}}{4x}=\underset{x\to \infty }{\text{lim}}\frac{6\text{ln}\hspace{0.17em}x}{4x^2}=\frac{\infty }{\infty }\text{???}$$ Try again: $$\underset{x\to \infty }{\text{lim}}\frac{6\text{ln}\hspace{0.17em}x}{4x^2}=\underset{x\to \infty }{\text{lim}}\frac{6\frac{1}{x}}{8x}=\underset{x\to \infty }{\text{lim}}\frac{6}{8x^2}=0\text{.}$$

Why does L'Hopital's rule work?

If $\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}\text{???}$ then $f\left(a\right)=0$ and $g\left(a\right)=0\text{.}$ Let $x=a+\mathrm{\Delta }x\text{.}$ Then $$\begin{array}{ccc}{\displaystyle \underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f(a+\mathrm{\Delta }x)}{g(a+\mathrm{\Delta }x)}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{g(a+\mathrm{\Delta }x)-g\left(a\right)}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{\mathrm{\Delta }x}}{\frac{g(a+\mathrm{\Delta }x)-g\left(a\right)}{\mathrm{\Delta }x}}}\\ & =& {\displaystyle \frac{\frac{df}{dx}{|}_{x=a}}{\frac{dg}{dx}{|}_{x=a}}}\\ & =& {\displaystyle \underset{x\to a}{\text{lim}}\frac{\frac{df}{dx}}{\frac{dg}{dx}}\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}}& =& {\displaystyle \underset{x\to a}{\text{lim}}\frac{\text{rate}\hspace{0.17em}f\left(x\right)\hspace{0.17em}\text{approaches 0}}{\text{rate}\hspace{0.17em}g\left(x\right)\hspace{0.17em}\text{approaches 0}}}\\ & =& {\displaystyle \underset{x\to a}{\text{lim}}\frac{\frac{df}{dx}}{\frac{dg}{dx}}\text{.}}\end{array}$$

Bad limits $\frac{0}{0}$ and $\frac{\infty }{\infty }$

$\underset{x\to 0}{\text{lim}}\frac{5x}{3x}=\frac{0}{0}\text{???}$ BAD $$\underset{x\to 0}{\text{lim}}=\frac{5x}{3x}=\underset{x\to 0}{\text{lim}}\frac{5}{3}=\frac{5}{3},$$ since $$\frac{\text{rate}\hspace{0.17em}5x\hspace{0.17em}\text{goes to}\hspace{0.17em}0}{\text{rate}\hspace{0.17em}3x\hspace{0.17em}\text{goes to}\hspace{0.17em}0}=\frac{\frac{d5x}{dx}}{\frac{d3x}{dx}}=\frac{5}{3},$$ i.e. $5x$ goes to $0$ $\frac{5}{3}$ as fast as $3x$ goes to $0\text{.}$

$\underset{x\to \infty }{\text{lim}}\frac{5x}{x}=\frac{\infty }{\infty }\text{???}$ BAD $$\underset{x\to \infty }{\text{lim}}\frac{5x}{x}$$ Here $5x$ goes to $\infty$ $5$ times as fast as $x$ goes to $\infty \text{.}$ $$\underset{x\to \infty }{\text{lim}}\frac{5x}{x}=\underset{x\to \infty }{\text{lim}}5=5\text{.}$$

$\underset{x\to 0}{\text{lim}}\frac{\frac{1}{x^2}}{\frac{1}{x^3}}=\frac{\infty }{\infty }\text{???}$ $$\underset{x\to 0}{\text{lim}}\frac{\frac{1}{x^2}}{\frac{1}{x^3}}=\underset{x\to 0}{\text{lim}}\frac{x^3}{x^2}=\underset{x\to 0}{\text{lim}}x=0\text{.}$$ Here $\underset{x\to 0}{\text{lim}}\frac{x^3}{x^2}=\frac{0}{0}\text{???}$ but $x^3$ goes to $0$ much faster than $x^2$ goes to $0\text{.}$ So $$\underset{x\to 0}{\text{lim}}\frac{x^3}{x^2}=0\text{.}$$

$\underset{x\to 0}{\text{lim}}\frac{-\text{ln}\hspace{0.17em}x}{\frac{1}{x}}=\frac{\infty }{\infty }\text{???}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{-\text{ln}\hspace{0.17em}x}{\frac{1}{x}}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{rate}\hspace{0.17em}-\text{ln}\hspace{0.17em}x\hspace{0.17em}\text{goes to}\hspace{0.17em}\infty }{\text{rate}\hspace{0.17em}\frac{1}{x}\hspace{0.17em}\text{goes to}\hspace{0.17em}\infty }}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{-\frac{1}{x}}{-\frac{1}{x^2}}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{x^2}{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}x}\\ & =& {\displaystyle 0\text{.}}\end{array}$$

Other bad limits $\infty -\infty ,$ $0·\infty ,$ $1^{\infty },$ $0^0$

(a) $0·\infty$

$\underset{x\to \pi }{\text{lim}}(x-\pi )\text{cot}\hspace{0.17em}x=0·\infty \text{???}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to \pi }{\text{lim}}(x-\pi )\text{cot}\hspace{0.17em}x}& =& {\displaystyle \underset{x\to \pi }{\text{lim}}\frac{x-\pi }{\text{tan}\hspace{0.17em}x}=\frac{0}{0}\text{???}}\\ {\displaystyle \underset{x\to \pi }{\text{lim}}\frac{x-\pi }{\text{tan}\hspace{0.17em}x}}& =& {\displaystyle \underset{x\to \pi }{\text{lim}}\frac{\text{rate that}\hspace{0.17em}x-\pi \hspace{0.17em}\text{goes to}\hspace{0.17em}0}{\text{rate that}\hspace{0.17em}\text{tan}\hspace{0.17em}x\hspace{0.17em}\text{goes to}\hspace{0.17em}0}}\\ & =& {\displaystyle \underset{x\to \pi }{\text{lim}}\frac{1}{{\text{sec}}^{2}x}=\underset{x\to \pi }{\text{lim}}{\text{cos}}^{2}x={\text{cos}}^2\pi ={(-1)}^2=1\text{.}}\end{array}$$

(b) $1^{\infty }$

$\underset{x\to 0}{\text{lim}}{(1-2x)}^{\frac{1}{x}}\stackrel{?}{=}{1}^{\infty }\text{???}$ BAD $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}{(1-2x)}^{\frac{1}{x}}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}{\left(e^{\text{ln}(1-2x)}\right)}^{\frac{1}{x}}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{\frac{1}{x}\text{ln}(1-2x)}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{\frac{-2\text{ln}(1-2x)}{-2x}}}\\ & =& {\displaystyle e^{-2·1}}\\ & =& {\displaystyle e^{-2}\text{.}}\end{array}$$

(c) $0^0$

$\underset{x\to 0}{\text{lim}}{x}^x=0^0\text{???}$ BAD $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}{x}^x}& =& {\displaystyle \underset{x\to 0}{\text{lim}}{\left(e^{\text{ln}\hspace{0.17em}x}\right)}^x}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{x\text{ln}\hspace{0.17em}x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{\frac{\text{ln}\hspace{0.17em}x}{\frac{1}{x}}}}\\ & =& {\displaystyle e^{\frac{\infty }{\infty }}\text{???}\text{BAD}}\\ {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{\frac{\text{ln}\hspace{0.17em}x}{\frac{1}{x}}}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{\frac{\frac{1}{x}}{-\frac{1}{x^2}}}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{\frac{-x^2}{x}}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}{e}^{-x}}\\ & =& {\displaystyle e^{-0}}\\ & =& {\displaystyle 1\text{.}}\end{array}$$

(d) $\infty -\infty$

$\underset{x\to 0}{\text{lim}}\frac{1}{x}-\frac{1}{x}=\infty -\infty \text{???}$ BAD $$\underset{x\to 0}{\text{lim}}\frac{1}{x}-\frac{1}{x}=\underset{x\to 0}{\text{lim}}0=0\text{.}$$

$\underset{x\to 0}{\text{lim}}{x}^{-1}-\text{csc}\hspace{0.17em}x=\infty -\infty \text{???}$ BAD $$\underset{x\to 0}{\text{lim}}{x}^{-1}-\text{csc}\hspace{0.17em}x=\underset{x\to 0}{\text{lim}}\frac{1}{x}-\frac{1}{\text{sin}\hspace{0.17em}x}=\underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}x-x}{x\text{sin}\hspace{0.17em}x}=\frac{0}{0}\text{???}$$ Try L'Hopital: $$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}x-x}{x\text{sin}\hspace{0.17em}x}=\underset{x\to 0}{\text{lim}}\frac{\text{cos}\hspace{0.17em}x-1}{\text{sin}\hspace{0.17em}x+x\text{cos}\hspace{0.17em}x}=\frac{0}{0}\text{???}$$ Try again: $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{cos}\hspace{0.17em}x-1}{\text{sin}\hspace{0.17em}x+x\text{cos}\hspace{0.17em}x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{cos}\hspace{0.17em}x-1}{\text{sin}\hspace{0.17em}x+x\text{cos}\hspace{0.17em}x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{-\text{sin}\hspace{0.17em}x}{\text{cos}\hspace{0.17em}x+\text{cos}\hspace{0.17em}x-x\text{sin}\hspace{0.17em}x}}\\ & =& {\displaystyle \frac{-0}{1+1-0}}\\ & =& {\displaystyle \frac{-0}{2}}\\ & =& {\displaystyle 0\text{.}}\end{array}$$

An example of when L'Hopital's rule doesn't work

$\underset{x\to 0}{\text{lim}}\frac{x}{-{\left(\text{ln}\hspace{0.17em}x\right)}^{-1}}=\frac{0}{0}\text{???}$ $$\underset{x\to 0}{\text{lim}}\frac{x}{-{\left(\text{ln}\hspace{0.17em}x\right)}^{-1}}=\underset{x\to 0}{\text{lim}}\frac{1}{+{\left(\text{ln}\hspace{0.17em}x\right)}^{-2}\frac{1}{x}}=\underset{x\to 0}{\text{lim}}\frac{x}{{\left(\text{ln}\hspace{0.17em}x\right)}^{-2}}=\frac{0}{0}\text{???}$$ Try again: $$\underset{x\to 0}{\text{lim}}\frac{x}{{\left(\text{ln}\hspace{0.17em}x\right)}^{-2}}=\underset{x\to 0}{\text{lim}}\frac{1}{-2{\left(\text{ln}\hspace{0.17em}x\right)}^{-3}\frac{1}{x}}=\underset{x\to 0}{\text{lim}}\frac{x}{-2{\left(\text{ln}\hspace{0.17em}x\right)}^{-3}}=\frac{0}{0}\text{???}$$ Try again: $$\underset{x\to 0}{\text{lim}}\frac{x}{-2{\left(\text{ln}\hspace{0.17em}x\right)}^{-3}}=\underset{x\to 0}{\text{lim}}\frac{1}{6{\left(\text{ln}\hspace{0.17em}x\right)}^{-4}\frac{1}{x}}=\underset{x\to 0}{\text{lim}}\frac{x}{6{\left(\text{ln}\hspace{0.17em}x\right)}^{-4}}=\frac{0}{0}\text{???}$$ $$\begin{array}{c}⋮\\ \text{Goes on forever}\end{array}$$ Instead do $$\underset{x\to 0}{\text{lim}}\frac{x}{-{\left(\text{ln}\hspace{0.17em}x\right)}^{-1}}=\underset{x\to 0}{\text{lim}}-x\text{ln}\hspace{0.17em}x=\underset{x\to 0}{\text{lim}}\frac{-\text{ln}\hspace{0.17em}x}{\frac{1}{x}}=\frac{\infty }{\infty }\text{???}$$ Then $$\underset{x\to 0}{\text{lim}}\frac{-\text{ln}\hspace{0.17em}x}{\frac{1}{x}}=\underset{x\to 0}{\text{lim}}\frac{-\frac{1}{x}}{-\frac{1}{x^2}}=\underset{x\to 0}{\text{lim}}\frac{x^2}{x}=\underset{x\to 0}{\text{lim}}x=0\text{.}$$

Notes and References

These are a typed copy of Lecture 36 from a series of handwritten lecture notes for the class MATH 221 given on December 6, 2000.

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