Continuity

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \DeclareMathOperator {\dx }{dx} \DeclareMathOperator {\dt }{dt} \DeclareMathOperator {\dy }{dy} \DeclareMathOperator {\dz }{dz} \DeclareMathOperator {\dr }{dr} \DeclareMathOperator {\dw }{dw} \DeclareMathOperator {\du }{du} \DeclareMathOperator {\dv }{dv} \DeclareMathOperator {\ds }{ds} \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \newcommand {\bar }[0]{\overline } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

The limit of a continuous function at a point is equal to the value of the function at that point.

Video Lecture

Explanation

Limits are simple to compute when they can be found by plugging the value into the function. That is, when $\lim _{x\rightarrow a}f(x) = f(a).$ We call this property continuity.

A function $f$ is continuous at a point $a$ if

(a): $\lim _{x\rightarrow a}f(x)$ exists.
(b): $f(a)$ exists (i.e. $a$ is in the domain of $f$ ).
(c): $\lim _{x\rightarrow a} f(x) = f(a)$ .

Often in literature you may see the alternative definition that $f$ is continuous at $a$ if $\lim \limits _{x\rightarrow a} f(x) = f(a)$ . This is actually the same thing, click the right blue arrow to the right if you want to know more!

This more compact form of “a function $f$ is continuous at a point $a$ ” is really making three statements:

(a): $f(a)$ is defined. That is, $a$ is in the domain of $f$ .
(b): $\lim _{x\rightarrow a} f(x)$ exists.
(c): $\lim _{x\rightarrow a} f(x) = f(a)$ .

Notice that $f(a)$ can’t fail to exist and still equal the limit of something else. In essence, the equality sign here requires both sides to exist before equality can be evaluated, thus writing just the last statement implies the first two.

Actual mathematicians love brevity and succinctness in their definitions; so they will commonly use the alternate definition as it is shorter and equivalent. But this brevity and succinctness that mathematicians value can often make it harder on students first learning a concept.

1 : Consider the graph of the function $f$

Which of the following are true?

$f$ is continuous at $x=0.5$ $f$ is continuous at $x=1$ $f$ is continuous at $x=1.5$

Find the discontinuities (the points $x$ where a function is not continuous) for the function described below:

To start, $f$ is not even defined at $x=\answer [given]{2}$ , hence $f$ cannot be continuous at $x=\answer [given]{2}$ .

Next, from the plot above we see that $\lim _{x\rightarrow 4} f(x)$ does not exist because $\lim _{x\rightarrow 4^-}f(x) = \answer [given]{1}\qquad \text {and}\qquad \lim _{x\rightarrow 4^+}f(x) \approx 3.3$ Since $\lim _{x\rightarrow 4} f(x)$ does not exist, $f$ cannot be continuous at $x=4$ .

We also see that $\lim _{x\rightarrow 6} f(x) \approx 2.9$ while $f(6) = \answer [given]{2}$ . Hence $\lim _{x\rightarrow 6} f(x) \ne f(6)$ , and so $f$ is not continuous at $x=6$ .

Building from the definition of continuity at a point, we can now define what it means for a function to be continuous on an open interval.

A function $f$ is continuous on an open interval $I = (b,c)$ if $\lim _{x\rightarrow a} f(x) = f(a)$ for all $a$ in $I$ .

Loosely speaking, a function is continuous on an open interval $I$ if you can draw the function on that interval without any breaks in the graph. This is often referred to as being able to draw the graph “without picking up your pencil.”

Continuity of Parent Functions The following functions are continuous on their natural domains, for $k$ a real number and $b$ a positive real number:

Constant function: $f(x) = k$
Identity function: $f(x) = x$
Power function: $f(x) = x^b$
Exponential function: $f(x) = b^x$
Logarithmic function: $f(x) = \log _b(x)$

In essence, we are saying that the functions listed above are continuous wherever they are defined.

2 : Compute: $\lim _{x\rightarrow \pi } x^3 = \answer {\pi ^3}$

The function $f(x)=x^3$ is of the form $x^b$ for a positive real number $b$ . Therefore, $f(x)=x^3$ is continuous for all positive real values of $x$ . In particular, $f(x)$ is continuous at $x=\pi$ . Since $x^3$ is continuous at $\pi$ , we know that $\lim _{x\rightarrow \pi } f(x) = f(\pi )$ . That is, $\lim _{x\rightarrow \pi } x^3 = \pi ^3$ .

3 : Compute: $\lim _{x\rightarrow \pi } \sqrt {2} = \answer {\sqrt {2}}$

The function $f(x)= \sqrt {2}$ is a constant. Therefore, $f(x)=\sqrt {2}$ is continuous for all real values of $x$ . In particular, $f(x)$ is continuous at $x=\pi$ . Since $f$ is continuous at $\pi$ , we know that $\lim _{x\rightarrow \pi } f(x) = f(\pi )=\sqrt {2}$ . That is, $\lim _{x\rightarrow \pi } \sqrt {2}= \sqrt {2}$ .

4 : Compute: $\lim _{x\rightarrow \pi } e^{x} = \answer {e^{\pi }}$

The function $f(x)= e^{x}$ is continuous for all real values of $x$ . In particular, $f(x)$ is continuous at $x=\pi$ . Since $f$ is continuous at $\pi$ , we know that $\lim _{x\rightarrow \pi } f(x) = f(\pi )=e^{\pi }$ . That is, $\lim _{x\rightarrow \pi } e^{x}= e^{\pi }$ .