Continuity

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For functions that are “normal” enough, we know immediately whetheror not they are continuous at a given point. Nevertheless, thecontinuity of a function is such an important property that we need aprecise definition of continuity at a point:

A function \$f\$ is continuous at \$c\$ if and only if \$displaystylelim_{xto c} f(x)=f(c).\$

That is, \$f\$ is continuous at \$c\$ if and only if for all \$varepsilon > 0\$there exists a \$delta > 0\$ such that [{smalltextrm{if }} x-c <delta quad{smalltextrm{then }} f(x)-f(c) <varepsilon.]In words, for \$x\$ close to \$c\$, \$f(x)\$ should be close to \$f(c)\$.

Notes

• If \$f\$ is continuous at every real number \$c\$, then \$f\$ is said to be continuous.
• If \$f\$ is not continuous at \$c\$, then \$f\$ is said to be discontinuous at \$c\$. The function \$f\$ can be discontinuous for two distinct reasons:
• \$f(x)\$ does not have a limit as \$xto c\$. (Specifically, if the left- and right-hand limits exist but are different, the discontinuity is called a jump discontinuity.)
• \$f(x)\$ has a limit as \$xto c\$, but \$lim_{xto c} f(x)neq f(c)\$ or \$f(c)\$ is undefined. (This is called a removable discontinuity, since we can “remove” the discontinuity at \$c\$ by redefining \$f(c)\$ as \$lim_{xto c} f(x)\$.)

\$f(x) = frac{x^2-9}{x-3}\$
Removable Discontinuity at \$x = 3\$

[ f(x)=left{begin{array}{ll} x^2, & x < 0 1, & x = 0x + 1, & x > 0end{array}right. ]
Discontinuity at \$x = 0\$

[ f(x)=left{begin{array}{ll} 1, & x < 0 3, & x leq 0end{array}right. ]
Jump Discontinuity at \$x = 0\$

Rather than returning to the \$varepsilon\$-\$delta\$ definitionwhenever we want to prove a function is continuous at a point, webuild up our collection of continuous functions by combining functionswe know are continuous:

If \$f\$ and \$g\$ are continuous at \$c\$, then

1. \$f+g\$ is continuous at \$c\$.
2. \$alpha f\$ is continuous at \$c\$ for any real number \$alpha\$.
3. \$fg\$ is continous at \$c\$.
4. \$f/g\$ is continuous at \$c\$ if \$g(c)neq 0\$.
Example

The function \$displaystyle f(x)=frac{x^2-4}{(x-2)(x-1)}\$ iscontinuous everywhere except at \$x=2\$ and at \$x=1\$. The discontinuityat \$x=2\$ is removable, since \$displaystyle frac{x^2-4}{(x-2)(x-1)}\$can be simplified to \$displaystyle frac{x+2}{x-1}\$. To remove thediscontinuity, define [f(2)=frac{2+2}{2-1}=4.]

We can also look at the composition \$fcirc g\$ of two functions,[(fcirc g)(x)=f(g(x)).]

If \$g\$ is continuous at \$c\$ and \$f\$ is continuous at \$g(c)\$, then the composition \$fcirc g\$ is continuous at \$c\$.

Theorem

Theorem

If \$g\$ is continuous at \$c\$ and \$f\$ is continuous at \$g(c)\$, then the composition \$fcirc g\$ is continuous at \$c\$.

Idea of Proof:

For \$x\$ “close” to \$c\$, \$g(x)\$ is “close” to \$g(c)\$.
For \$g(x)\$ “close” to \$g(c)\$, \$f(g(x))\$ is “close” to \$f(g(c))\$.
So for \$x\$ “close” to \$c\$, \$f(g(x))\$ is “close” to \$f(g(c))\$.

Proof:

Let \$varepsilon>0\$.

We will show that there exists \$delta>0\$ such that if \$ x-c <delta\$, then \$ f(g(x))-f(g(c)) <varepsilon\$.

Since \$f\$ is continuous at \$g(c)\$, there exists \$delta_1>0\$ such that

\$\$ {smalltextrm{if }} x_1-g(c) <delta_1, {smalltextrm{ then }} f(x_1)-f(g(c)) <varepsilon. \$\$For \$delta_1>0\$, there exists \$delta>0\$ such that \$\$ {smalltextrm{if }} x-c <delta, {smalltextrm{ then }} g(x)-g(c) <delta_1 \$\$by the continuity of \$g\$ at \$c\$.

Substituting \$g(x)\$ for \$x_1\$ in (1), we have from (2)and (1) that[{smalltextrm{if }} x-c <delta, {smalltextrm{ then }} f(g(x))-f(g(c)) <varepsilon,]and the proof is complete.

Continuity Math

(This proof is taken from Salas and Hille’s Calculus: OneVariable, 7th ed.)

We’d also like to speak of continuity on a closed interval \$[a,b]\$.To deal with the endpoints \$a\$ and \$b\$, we define one-sided continuity:

A function \$f\$ is continuous from the left at \$c\$ if and onlyif \$displaystylelim_{xto c^-} f(x)=f(c)\$. It is continuous from theright at \$c\$ if and only if \$displaystylelim_{xto c^+} f(x)=f(c)\$.

We say that \$f\$ is continuous on \$[a,b]\$ if and only if

1. \$f\$ is continuous on \$(a,b)\$,
2. \$f\$ is continuous from the right at \$a\$, and
3. \$f\$ is continuous from the left at \$b\$.
One-Sided Continuity

\$displaystyle{lim_{xrightarrow 0^+} f(x)=f(0)}\$ so \$f(x) = sqrt{x}\$ is continuous from the right at \$0\$.

\$displaystyle{lim_{xrightarrow 1^-} f(x)=f(1)}\$ so \$f(x)=sqrt{1-x}\$ is continuous from the left at \$1\$.

Note that \$f\$ is continuous at \$c\$ if and only if the right- andleft-hand limits exist and both equal \$f(c)\$.

Example

The function [ f(x)=left{begin{array}{ll} x, & xleq 0 x^2, & 0 < xleq 1 frac{2}{x}, & 1 < xleq 2 x-1, & x > 2 end{array}right. ] is continuous everywhere except at \$x=1\$, where \$f\$ has a jump discontinuity.

Continuity Tester Key Concepts

A function \$f\$ is continuous at \$c\$ if and only if \$displaystylelim_{x to c} f(x) = f(c)\$.

That is, \$f\$ is continuous at \$c\$ if and only if for all \$varepsilon > 0\$ there exists a \$delta > 0\$ such that[{smalltextrm{if }} x-c < delta {smalltextrm{ then }} f(x)-f(c) < varepsilon{smalltextrm{.}}]In words, for \$x\$ close to \$c\$, \$f(x)\$ should be close to \$f(c)\$.

Continuity Tester

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