Tutorial: Limits: algebraic viewpoint
Adaptive game version
This tutorial: Part B: Continuity from the algebraic viewpoint
(This topic is also in Section 10.3 in Finite Mathematics and Applied Calculus)
Resources
 Function evaluator and grapher |
True false quiz |
Continuities and discontinuities algebraically
Let us look once again at the definition of continuity we saw in %%partAtut:
#[Continuous function][Función continua]#
Let $x = a$ be a point in the domain of the function $f.$ Then $f$ is %%contat $x = a$ if both of the following are true:
1. $\displaystyle \lim_{x \to a} f(x)$ %%exists.\t $\qquad$ \t
So, if the left and right limits exist, they are equal.
\\ 2. $\displaystyle \lim_{x \to a} f(x) = f(a)$ \t \t
The limit equals the value of the function at the point $a$.
The function $f$ is continuous on its domain if it is continuous at every point in its domain.
Theorem C: Limits of closed form functions
If $f$ is a closed-form function and $f(a)$ is defined, then $\displaystyle \lim_{x \to a} f(x)$ exists, and equals $f(a).$
#[In other words,][En otras palabras]#,
Corollary C: Continuity of closed form functions
If $f$ is a closed-form function $a$ is in the domain of $f$, then $f$ is continuous at $a$. Thus, closed form functions are continuous on their domains!
Examples
- $f(x) = \dfrac{2x^2+7x+3}{x + 3}$ is a closed-form function. and $2$ is in the domain of $f$. So, $f$ is continuous at $2$.
- Because $-1$ is also in the domain of $f$, $f$ is continuous at $-1$ as well.
$\displaystyle \lim_{x \to -1} \frac{2x^2+7x+3}{x+3} = f(-1) -\frac{-2}{2} = -1$.
- However, $-3$ is not in the domain of $f$, and so we cannot say that $f$ is continuous at $-3$; in fact it is singular there.
Functions not in closed form
Even when a function is not given in closed form but instead is piecewise defined, we can still analyze its continuity algebraically (in %%partAtut we looked at such functions from the graphical viepoint).
Let $f$ be the function given by
$f(x) = \begin{cases} -1 & \text{ if } -4 \leq x \lt -1 \\ x & \text{ if } -1 \leq x \leq 1 \\ x^2-1 & \text { if } 1 \lt x \leq 2 \end{cases}$
As in the preceding tutorial, we are going to look at the continuity of $f$ at various points, but this time purely algebraically and without even looking at the graph. You can see the graph by clicking here if you really want to, but try not to look.
#[Example][Ejemplo]#
#[Determine whether the above function $f$ is continuous at following points:][Determina si o no la función $f$ de arriba es continua en los siguientes puntos:]#
a. $x = 0$ \gap[10] b. $x = -1$ \gap[10] c. $x = 1$ \gap[10] d. $x = 2$
#[Solution][>Solución]#
Whether or not $f$ is continuous at $x = a$ depends on the limit as $x \to a$, and that limit depends only on the values of $f$ close to, and on either side of $x = a$ (if $a$ is not an endpoint of the domain). Put another way:
a. $x = 0$ is in the interval $[-1, 1]$, on which $f(x) = x$. Moreover, $x = 0$ is an interior point (not an endpoint) of that interval, so that there is an open interval about $x = 0$ (for instance, $(-1/2, 1/2)$) on which $f(x) = x$, a closed-form function. Therefore, $f$ is continuous at $x = 0$. b. $x = -1$ is also in the interval $[-1 , 1]$. However, $x = -1$ is not an interior point; every open interval about $x = -1$ also contains points in the first interval $[-4, -1)$, so we cannot say that $f$ is closed-form on some open interval about $x = -1$. Instead, we calculate the left and right limits separately:
$\qquad \quad f(x) = \dfrac{2x^2+7x+3}{x+3}$
$\qquad \quad$ #[Singular at][Singular en]# $x = -3$
$\qquad \qquad f(x) = 2x+1$
$\qquad \quad$#[Continuous at][Continua en]# $x = -3$
Last Updated: August 2022
Copyright © 2022 Stefan Waner and Steven R. Costenoble
Whether $f$ is continuous at $x = a$ depends only on what $f$ looks like in some open interval about $x = a$.
a. $x = 0$ is in the interval $[-1, 1]$, on which $f(x) = x$. Moreover, $x = 0$ is an interior point (not an endpoint) of that interval, so that there is an open interval about $x = 0$ (for instance, $(-1/2, 1/2)$) on which $f(x) = x$, a closed-form function. Therefore, $f$ is continuous at $x = 0$. b. $x = -1$ is also in the interval $[-1 , 1]$. However, $x = -1$ is not an interior point; every open interval about $x = -1$ also contains points in the first interval $[-4, -1)$, so we cannot say that $f$ is closed-form on some open interval about $x = -1$. Instead, we calculate the left and right limits separately:
$\displaystyle \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-1) = -1 \qquad$ \t #[because][porque]# $f(x) = -1$ #[for][para]# $x \lt -1$
\\ $\displaystyle \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} x = -1$ \t #[because][porque]# $f(x) = x$ #[for][para]# $x \gt -1$
Thus, $\displaystyle \lim_{x \to -1} f(x)$ exists and equals $-1$. Also, from the formula, $f(-1) = -1$. Since the limit exists and equals $f(-1)$, the function $f$ is continuous at $x = -1$.
c.
$x = 1$ is the other non-interior point of $[-1 , 1]$; every open interval about $x = 1$ also contains points in the third interval $(1, 2]$, so again we cannot say that $f$ is closed-form on some open interval about $x = 1$, and we must again calculate the left and right limits:
$\displaystyle \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$ \t #[because][porque]# $f(x) = x$ #[for][para]# $x \lt 1$
\\ $\displaystyle \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 - 1 = 0 \qquad$ \t #[because][porque]# $f(x) = x^2 - 1$ #[for][para]# $x \gt 1$
Thus, $\displaystyle \lim_{x \to 1} f(x)$ does not exist, and so the function $f$ is discontinuous at $x = -1$
d.
Although $x = 2$ is not an interior point of its interval $(1, 2]$, it is the right endpoint of the domain of $f$, so the limit is equal to the left limit:
$\displaystyle \lim_{x \to 2} f(x) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 - 1 = 3 = f(2)$.
Thus, $f$ is continuous at 2.
Removable and essential singularities
In %%limitalgebraicallytutA we considered the limit
$\displaystyle \lim_{x \to -3} \frac{2x^2+7x+3}{x+3}$.
#[using a simplification that removed the singularity at $x = -3$][usando una simplificación que eliminó la singularidad a $x = -3$]#:
$\dfrac{2x^2+7x+3}{x+3}$ \t ${}= \dfrac{(2x+1)\color{blue}{(x+3)}}{\color{blue}{x+3}}$ \t #[Singular at][Singular en]# $x=-3$
\\ \t ${}= 2x+1 \quad (x \ne -3)$ \t #[Not singular at][Nosingular en]# $x=-3$
#[We changed the function to a new one that is continuous at $x = -3$, and no longer singular there. For this reason, we refer to the original singularity as removable.][Cambiamos la función a una nueva que e continua en $x=-3$, y ya no es singular allí. Por esta razón, nos referimos a la singularidad original como removible.]#
$\qquad \quad f(x) = \dfrac{2x^2+7x+3}{x+3}$
$\qquad \quad$ #[Singular at][Singular en]# $x = -3$
$\qquad \qquad f(x) = 2x+1$
$\qquad \quad$#[Continuous at][Continua en]# $x = -3$
Now try the exercises in Section 10.3 in Finite Mathematics and Applied Calculus.
or move ahead to the next part of this tutorial by pressing "Next" on the sidebar.
Copyright © 2022 Stefan Waner and Steven R. Costenoble