MATH 221

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

Last updated: 13 August 2014

Lecture 13

Existence of limits

What is $\underset{x\to 0}{\text{lim}}\frac{1}{x}\text{?}$ If $x=.1$ then $\frac{1}{x}=10\text{.}$
If $x=.01$ then $\frac{1}{x}=100\text{.}$
If $x=.001$ then $\frac{1}{x}=100\text{.}$
If $x=.0001$ then $\frac{1}{x}=100\text{.}$ So, it looks like $\underset{x\to 0}{\text{lim}}\frac{1}{x}=\infty \text{.}$
If $x=-.1$ then $\frac{1}{x}=-10\text{.}$
If $x=-.01$ then $\frac{1}{x}=-100\text{.}$
If $x=-.001$ then $\frac{1}{x}=-100\text{.}$
If $x=-.0001$ then $\frac{1}{x}=-100\text{.}$ So, it looks like $\underset{x\to 0}{\text{lim}}\frac{1}{x}=-\infty \text{.}$
Since $\underset{x\to 0^{+}}{\text{lim}}\frac{1}{x}=\infty$ and $\underset{x\to 0^{-}}{\text{lim}}\frac{1}{x}=\infty ,$ $\underset{x\to 0}{\text{lim}}\frac{1}{x}=\text{UNDEFINED.}$ $$\begin{array}{c} y x -1 1 -1 1 f⁡(x)=1x \end{array}$$

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

Look at the graph of $\text{ln}\hspace{0.17em}x\text{.}$

So, from the graph, $\text{ln}\hspace{0.17em}x$ doesn't even make sense for $x$ close to $-1\text{.}$ So $\underset{x\to -1}{\text{lim}}\text{ln}\hspace{0.17em}x$ is certainly undefined.

Note: If we allow $x$ to get closer and closer to $-1$ and be a complex number then $$\text{ln}\hspace{0.17em}-1=i\pi \text{and}i3\pi \text{and}i5\pi \dots$$ since $e^{i\pi }=\text{cos}\hspace{0.17em}\pi +i\text{sin}\hspace{0.17em}\pi =-1+i·0=-1$ and $\text{ln}\hspace{0.17em}-1=i\pi \text{.}$ Still $\underset{x\to -1}{\text{lim}}\text{ln}\hspace{0.17em}x$ is undefined since it can't be $i\pi$ and $3i\pi$ and $5i\pi \dots$ all at once.

$\underset{x\to \infty }{\text{lim}}\text{sin}\hspace{0.17em}x$

The graph of $\text{sin}\hspace{0.17em}x$ is $$\begin{array}{c} x y 1 -1 y=sin⁡x \end{array}$$ So, as $x$ gets larger and larger, $\text{sin}\hspace{0.17em}x$ keeps going back and forth between $-1$ and $+1\text{.}$ So $\text{sin}\hspace{0.17em}x$ doesn't get closer and closer to anything as $x$ gets larger and larger. So $\underset{x\to \infty }{\text{lim}}\text{sin}\hspace{0.17em}x$ is undefined.

Continuous functions

A function is continuous if $f\left(x\right)$ doesn't jump when $x$ changes. The function $f\left(x\right)$ is not continuous exactly at the places where it jumps.

A function $f\left(x\right)$ is continuous at $x=a$ if it doesn't jump at $x=a,$ i.e. if $\underset{x\to a}{\text{lim}}f\left(x\right)=f\left(a\right)\text{.}$ $$\begin{array}{c} a x y y=f⁡(x) \end{array}$$ Not continuous at $x=a\text{.}$

$f\left(x\right)=\lfloor x\rfloor$ is the round down function. $\lfloor 3.2\rfloor =3\text{.}$

$$\begin{array}{c} -5 -4 -3 -2 1 2 3 4 5 x -4 -3 -2 -1 1 2 3 4 y \end{array}$$ $f\left(x\right)$ is continuous if $x\ne 0,\pm 1,\pm 2,\pm 3,\dots$

Note: $\underset{x\to 1^{-}}{\text{lim}}\lfloor x\rfloor =0$ and $\underset{x\to 1^{+}}{\text{lim}}\lfloor x\rfloor =1\text{.}$

$f\left(x\right)=\lceil x\rceil$ is the round up function. $\lceil 3.2\rceil =4\text{.}$

$f\left(x\right)=\{\begin{array}{cc}1+x^2,& 0\le x\le 1,\\ 2-x,& x>1\text{.}\end{array}$ $$\begin{array}{c} x y 1 2 3 1 2 y=f⁡(x) \end{array}$$ $f\left(x\right)$ jumps at $x=1\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 1^{-}}{\text{lim}}f\left(x\right)}& =& {\displaystyle \underset{x\to 1^{-}}{\text{lim}}1+x^2=2\text{.}}\\ {\displaystyle \underset{x\to 1^{+}}{\text{lim}}f\left(a\right)}& =& {\displaystyle \underset{x\to 1^{+}}{\text{lim}}2-x=1\text{.}}\end{array}$$ So $\underset{x\to 1}{\text{lim}}f\left(x\right)$ is UNDEFINED.

$$f\left(x\right)=\{\begin{array}{cc}\frac{\text{sin}\hspace{0.17em}3x}{x},& \text{if}\hspace{0.17em}x\ne 0,\\ 1,& \text{if}\hspace{0.17em}x=0\text{.}\end{array}$$ $\text{sin}\hspace{0.17em}3x$ is continuous everywhere and $x$ is continuous everywhere. So $\frac{\text{sin}\hspace{0.17em}3x}{x}$ is continuous everywhere. EXCEPT, it makes no sense when $x=0\text{.}$ Now what is happening when $x=0\text{?}$ $$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}3x}{x}=\underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}3x}{3x}·3=1·3=3\text{.}$$ BUT $$f\left(0\right)=1,$$ So $$\underset{x\to 0}{\text{lim}}f\left(x\right)\ne f\left(0\right)$$ in this case. So $f\left(x\right)$ is not continuous when $x=0\text{.}$ $$\begin{array}{c} y 1 -1 x -3π -5π2 -2π -3π2 -π -π2 π2 π 3π2 2π 5π2 3π y=sin⁡x \\ y 1 -1 x -3π -5π2 -2π -3π2 -π -π2 π2 π 3π2 2π 5π2 3π y=sin⁡3x \\ x y y=1x x y y=sin⁡3xx \end{array}$$

Notes and References

These are a typed copy of Lecture 13 from a series of handwritten lecture notes for the class MATH 221 given on October 6, 2000.

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