Continuity Markup: Use your iPad, iPhone, or iPod touch to add sketches, shapes, and other markup to a Mac document, and see the changes live on your Mac. Sidecar: Use your iPad as a second display that extends or mirrors your Mac desktop. Continuity (fiction), consistency of plot elements, such as characterization, location, and costuming, within a work of fiction (this is a mass noun) Continuity (setting), one of several similar but distinct fictional universes in a broad franchise of related works (this is a count noun).

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10 synonyms of continuity from the Merriam-Webster Thesaurus, plus 10 related words, definitions, and antonyms. Find another word for continuity. Continuity: uninterrupted or lasting existence. Synonyms: abidance, ceaselessness, continuance Antonyms: cessation, close, discontinuance Find the right word.

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 $f(c)$.


  • 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 $g(c)neq 0$.

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 $c$.



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 $f(g(c))$.


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 $f(c)$.


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.

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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 $f(c)$.

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