# 3.3 Continuity  (Page 2/2)

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Finally, using the continuity of both $f$ and $g$ applied to the positive numbers ${ϵ}_{1}=ϵ/\left(4{M}_{2}{|g\left(c\right)|}^{2}\right)$ and ${ϵ}_{2}=ϵ/\left(4{M}_{1}{|g\left(c\right)|}^{2}\right),$ choose $\delta >0,$ with $\delta and such that if $|y-c|<\delta$ and $y\in S$ then $|f\left(y\right)-f\left(c\right)|<\frac{ϵ}{4{M}_{2}/{|g\left(c\right)|}^{2}}$ and $|g\left(c\right)-g\left(y\right)|<\frac{ϵ}{4{M}_{1}/{|g\left(c\right)|}^{2}}.$ Then, if $|y-c|<\delta$ and $y\in S$ we have that

$|\frac{f\left(y\right)}{g\left(y\right)}-\frac{f\left(c\right)}{g\left(c\right)}|<ϵ$

as desired.

1. Prove part (2) of the preceding theorem. (It's an $ϵ/2$ argument.)
2. Prove part (3) of the preceding theorem. (It's similar to the proof of part (5) only easier.)
3. Prove part (4) of the preceding theorem.
4. Prove part (6) of the preceding theorem.
5. Suppose $S$ is a subset of $R.$ Verify the above theoremreplacing “ continuity” with left continuity and right continuity.
6. If $S$ is a subset of $R,$ show that $f$ is continuous at a point $c\in S$ if and only if it is both right continuous and left continuous at $c.$

## The composition of continuous functions is continuous.

Let $S,T,$ and $U$ be subsets of $C,$ and let $f:S\to T$ and $g:T\to U$ be functions. Suppose $f$ is continuous at a point $c\in S$ and that $g$ is continuous at the point $f\left(c\right)\in T.$ Then the composition $g\circ f$ is continuous at $c.$

Let $ϵ>0$ be given. Because $g$ is continuous at the point $f\left(c\right),$ there exists an $\alpha >0$ such that $|g\left(t\right)-g\left(f\left(c\right)\right)|<ϵ$ if $|t-f\left(c\right)|<\alpha .$ Now, using this positive number $\alpha ,$ and using the fact that $f$ is continuous at the point $c,$ there exists a $\delta >0$ so that $|f\left(s\right)-f\left(c\right)|<\alpha$ if $|s-c|<\delta .$ Therefore, if $|s-c|<\delta ,$ then $|f\left(s\right)-f\left(c\right)|<\alpha ,$ and hence $|g\left(f\left(s\right)\right)-g\left(f\left(c\right)\right)|=|g\circ f\left(s\right)-g\circ f\left(c\right)|<ϵ,$ which completes the proof.

1. If $f:C\to C$ is the function defined by $f\left(z\right)=z,$ prove that $f$ is continuous at each point of $C.$
2. Use part (a) and Theorem 3.2 to conclude that every rational function is continuous on its domain.
3. Prove that a step function $h:\left[a,b\right]\to C$ is continuous everywhere on $\left[a,b\right]$ except possibly at the points of the partition $P$ that determines $h.$
1. Let $S$ be the set of nonnegative real numbers, and define $f:S\to S$ by $f\left(x\right)=\sqrt{x}.$ Prove that $f$ is continuous at each point of $S.$ HINT: For $c=0,$ use $\delta ={ϵ}^{2}.$ For $c\ne 0,$ use the identity
$\sqrt{y}-\sqrt{c}=\left(\sqrt{y}-\sqrt{c}\right)\frac{\sqrt{y}+\sqrt{c}}{\sqrt{y}+\sqrt{c}}=\frac{y-c}{\sqrt{y}+\sqrt{c}}\le \frac{y-c}{\sqrt{c}}.$
2. If $f:C\to R$ is the function defined by $f\left(z\right)=|z|,$ show that $f$ is continuous at every point of its domain.

Using the previous theorems and exercises, explain why the following functions $f$ are continuous on their domains. Describe the domains as well.

1. $f\left(z\right)=\left(1-{z}^{2}\right)/\left(1+{z}^{2}\right).$
2. $f\left(z\right)=|1+z+{z}^{2}+{z}^{3}-\left(1/z\right)|.$
3. $f\left(z\right)=\sqrt{1+\sqrt{1-{|z|}^{2}}}.$
1. If $c$ and $d$ are real numbers, show that $max\left(c,d\right)=\left(c+d\right)/2+|c-d|/2.$
2. If $f$ and $g$ are functions from $S$ into $R,$ show that $max\left(f,g\right)=\left(f+g\right)/2+|f-g|/2.$
3. If $f$ and $g$ are real-valued functions that are both continuous at a point $c,$ show that $max\left(f,g\right)$ and $min\left(f,g\right)$ are both continuous at $c.$

Let $N$ be the set of natural numbers, let $P$ be the set of positive real numbers, and define $f:N\to P$ by $f\left(n\right)=\sqrt{1+n}.$ Prove that $f$ is continuous at each point of $N.$ Show in fact that every function $f:N\to C$ is continuous on this domain $N.$

HINT: Show that for any $ϵ>0,$ the choice of $\delta =1$ will work.

1. Negate the statement: “For every $ϵ>0,$ $|x|<ϵ{.}^{\text{'}\text{'}}$
2. Negate the statement: “For every $ϵ>0,$ there exists an $x$ for which $|x|<ϵ{.}^{\text{'}\text{'}}$
3. Negate the statement that “ $f$ is continuous at $c{.}^{\text{'}\text{'}}$

The next result establishes an equivalence between the basic $ϵ,\delta$ definition of continuity and a sequential formulation.In many cases, maybe most, this sequential version of continuity is easier to work with than the $ϵ,\delta$ version.

Let $f:S\to C$ be a complex-valued function on $S,$ and let $c$ be a point in $S.$ Then $f$ is continuous at $c$ if and only if the following condition holds: For every sequence $\left\{{x}_{n}\right\}$ of elements of $S$ that converges to $c,$ the sequence $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left(c\right).$ Or, said a different way, if $\left\{{x}_{n}\right\}$ converges to $c,$ then $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left(c\right).$ And, said yet a third (somewhat less precise) way, the function $f$ converts convergent sequences to convergent sequences.

Suppose first that $f$ is continuous at $c,$ and let $\left\{{x}_{n}\right\}$ be a sequence of elements of $S$ that converges to $c.$ Let $ϵ>0$ be given. We must find a natural number $N$ such that if $n\ge N$ then $|f\left({x}_{n}\right)-f\left(c\right)|<ϵ.$ First, choose $\delta >0$ so that $|f\left(y\right)-f\left(c\right)|<ϵ$ whenever $y\in S$ and $|y-c|<\delta .$ Now, choose $N$ so that $|{x}_{n}-c|<\delta$ whenever $n\ge N.$ Then if $n\ge N,$ we have that $|{x}_{n}-c|<\delta ,$ whence $|f\left({x}_{n}\right)-f\left(c\right)|<ϵ.$ This shows that the sequence $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left(c\right),$ as desired.

We prove the converse by proving the contrapositive statement; i.e., we will show that if $f$ is not continuous at $c,$ then there does exist a sequence $\left\{{x}_{n}\right\}$ that converges to $c$ but for which the sequence $\left\{f\left({x}_{n}\right)\right\}$ does not converge to $f\left(c\right).$ Thus, suppose $f$ is not continuous at $c.$ Then there exists an ${ϵ}_{0}>0$ such that for every $\delta >0$ there is a $y\in S$ such that $|y-c|<\delta$ but $|f\left(y\right)-f\left(c\right)|\ge {ϵ}_{0}.$ To obtain a sequence, we apply this statement to $\delta$ 's of the form $\delta =1/n.$ Hence, for every natural number $n$ there exists a point ${x}_{n}\in S$ such that $|{x}_{n}-c|<1/n$ but $|f\left({x}_{n}\right)-f\left(c\right)|\ge {ϵ}_{0}.$ Clearly, the sequence $\left\{{x}_{n}\right\}$ converges to $c$ since $|{x}_{n}-c|<1/n.$ On the other hand, the sequence $\left\{f\left({x}_{n}\right)\right\}$ cannot be converging to $f\left(c\right),$ because $|f\left({x}_{n}\right)-f\left(c\right)|$ is always $\ge {ϵ}_{0}.$

This completes the proof of the theorem.

how do they get the third part x = (32)5/4
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Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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