Continuity exercises

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Here is an opportunity for you to practice using the definition of continuity.

Consider the function $f(x) = \frac {x+4}{x^2-16}.$ Is this function continuous on $\RR = (\infty , \infty )?$

Yes No

Now that you know this function is not continuous everywhere, select all of the $x$ values at which this function is not continuous.

$x = 4$ $x=-16$ $x=16$ $x=-4$ $x=0$

You got it - $f$ is discontinuous at $x=-4$ and $x=-4$ . Let’s use limits and function values to determine what type of discontinuity $f$ has at $x=-4$ .

First, make a prediction. Any prediction you make is correct because it’s what you think currently, so take a guess! At $x=-4$ , I predict that $f$ has a(n)

removable discontinuity/holejump discontinuityinfinite discontinuityoscillating discontinuity .

Now, evaluate the following in order to make a conclusion about the type of discontinuity $f$ has at $x=-4$ . If a limit or function value does not exist, write DNE.

$\displaystyle \lim _{x \to -4} f(x) = \answer {\frac {-1}{8}}$
$f(-4) = \answer {DNE}$

Based on this, you can conclude that at $x=-4$ $f$ has a(n)

removable discontinuity/holejump discontinuityinfinite discontinuityoscillating discontinuity .

Consider the function $g(x) = 5x^4 - \pi x^2 - 5.$ Is this function continuous on $\RR = (\infty , \infty )?$

Yes No

Absolutely. $g(x)$ is a polynomial, and polynomials are continuous on $(-\infty , \infty )$ .

Consider the piece-wise function

$h(x) = \begin {cases} 3x+5 & x<0 \\ x^2 & x \geq 0 \end {cases}$

Is this function continuous on $\RR = (\infty , \infty )?$

Yes No

You got it: if you look closely, this function is discontinuous at $x=0$ . Select the reason why this function is discontinuous at $x=0$ below.

$h(0)$ does not exist $\displaystyle \lim _{x \to 0} h(x)$ does not exist $h(0) \neq \displaystyle \lim _{x \to 0} h(x)$

The main reason why $h(x)$ is discontinuous at $x=0$ is because $\displaystyle \lim _{x \to 0} h(x)$ does not exist. You could also say that $h(0) \neq \displaystyle \lim _{x \to 0} h(x)$ because $h(0)$ exists and $\displaystyle \lim _{x \to 0} h(x)$ doesn’t.

Consider the piece-wise function

$j(x) = \begin {cases} x^3 & x < 1 \\ 5 & x=1 \\ 2x-1 & x > 1 \end {cases}$

Is this function continuous on $\RR = (\infty , \infty )?$

Yes No

Good thinking: this function is discontinuous at $x=1$ . Select the reason why this function is discontinuous at $x=1$ below.

$j(1)$ does not exist $\displaystyle \lim _{x \to 1} j(x)$ does not exist $j(1) \neq \displaystyle \lim _{x \to 1} j(x)$

Let’s change $j(x)$ just slightly:

$j(x) = \begin {cases} x^3 & x < 1 \\ 1 & x=1 \\ 2x-1 & x > 1 \end {cases}$

With this change, is this $j(x)$ continuous on $\RR = (\infty , \infty )?$

Yes No

You’ll notice that changing a single number resulted in $j(x)$ being continuous on $\RR$ because now $j(1) = \displaystyle \lim _{x \to 1} j(x)$ . The small details really do matter!

Let

$f(x) = \begin {cases} 3x+1 & x < 1 \\ \sqrt {A+2x} & x \geq 1 \end {cases}$

where $A$ is a constant.

When $A =\answer {14}$ , $f(x)$ is continuous everywhere.

Let

$z(x) = \begin {cases} Bx-2 & x < 4 \\ Cx & x = 4 \\ -\frac {1}{2} Bx +10 & x > 4 \end {cases}$

where $B$ and $C$ are constants.

When $B =\answer {2}$ and $C = \answer {\frac {3}{2}}$ , $z(x)$ is continuous everywhere.

Consider the statement below, and then indicate whether it is sometimes, always, or never true.

‘‘If $f$ is continuous at $x=-2$ , then $\displaystyle \lim _{x\to -2} f(x)$ exists.”

This statement is

sometimesalwaysnever true.

You may want to review the definition of continuity on this card if you are struggling with this question.

Consider the statement below, and then indicate whether it is sometimes, always, or never true.

‘‘If $\displaystyle \lim _{x\to -2} f(x)$ exists, then $f$ is continuous at $x=-2$ .”

This statement is

sometimesalwaysnever true.

Now that you know this statement is sometimes true, try to draw an example of a graph for which it is true and an example of a graph for which it is false.

Consider the statement below, and then indicate whether it is sometimes, always, or never true.

‘‘If $f$ is a continuous function on $(-5, 5)$ , $f(-5) = -3$ , and $f(5) = 3$ , then there is an $x$ -value $c$ in $(-5,5)$ such that $f(c) = 0$ .”

This statement is

sometimesalwaysnever true.

At first glance, this seems like an example of the Intermediate Value Theorem which would always be true; however, there’s a small (but important) difference between this statement and the IVT. Can you spot it?