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  1. Math 242: Calculus 1/
  2. Chapter 1/

Section 1.8


Definition of continuous at a point #

A function $f$ is continuous at a number $a$ if $\displaystyle \lim_{x\to a} f(x) = f(a)$.

That definition tells us a lot! #

Writing $\displaystyle \lim_{x\to a} f(x) = f(a)$ means the following three things:

  1. There is a point $( a, f(a) )$ on the graph; that is, $f(a)$ exists.
  2. $\displaystyle \lim_{x\to a} f(x)$ exists
  3. $\displaystyle \lim_{x\to a} f(x) = f(a)$ , that is, we can just plug in $a$ to find the limit!

Definition: Discontinuous at a point #

If $f$ is defined near $a$ but not defined at $x=a$, then we say that $f$ is discontinuous at $x=a$. Or, we say $f$ has a discontinuity at $x=a$.

Example #

Identify where the function is discontinuous:


Example #

Identify where the function is discontinuous:

  1. $f(x) = \dfrac{3x^2 + 2x+1}{x-2}$

    Click to reveal the answer. Here the function is discontinuous at $x=2$. This is a jump type discontinuity (the graph jumps over the asymptote) It is continuous everywhere else.
  2. $g(x) = \dfrac{1}{x^2}$

    Click to reveal the answer. Like above, the function is discontinuous at $x=0$. This is a jump type discontinuity (the graph jumps over the asymptote) It is continuous everywhere else.
  3. $h(x) = \left\{\begin{align} \dfrac{x^2-4}{x-2} & \text{, if } x\ne 2 \\ 4 &\text{, if } x=2 \end{align} \right.$

    Hint: find a limit

    Click to reveal the answer. This one's neat! Since $\lim_{x\to 2} = 4$ (we did example in the [video in section 1.6]( and $f(2)=4$, the limit exists and equals the value of the function. It's continuous everywhere since we plugged the hole!

Continuity on an interval #

Note We define

  • continuity from the left at a if $\displaystyle \lim_{x\to a^-} f(x) = f(a)$, and
  • continuity from the right at a if $\displaystyle \lim_{x\to a^+} f(x) = f(a)$

Definition #

A function $f$ is said to be continuous on an interval $(a, b)$ if it continous for every number in the interval in $(a,b)$.

Example #

Revisiting our function from before, on what intervals is $f$ continuous?

Click to reveal the answer. $f$ is continuous on the intervals $(-\infty, -5)$, $(-5, 2)$, $(2, 7)$, and $(7, \infty)$. We can write that with union notation[^1]: $(-\infty, -5) \cup (-5, 2)\cup(2, 7)\cup(7, \infty)$.

Facts about continuity #

  1. Polynomials are continuous everywhere - all $x$ values in $(\infty, \infty)$.
  2. A rational function is continuous on its domain
  3. Root functions, trig functions and exponentials and logarithms1 are continuous on their domains.

If you still haven’t memorized the domains of trig functions, this would be a good time to take a moment to start. Especially the sine, cosine, and tangent functions.

Example #

Show that the piecewise function $$ f(x) = \left\{ \begin{align} x &\text{, if } x < 1 \\ \sqrt{x} &\text{, if } x \ge 1 \end{align} \right. $$ is continuous on $(\infty, \infty)$.

Example #

Please attempt the following example before checking the answer:

On what intervals are the following continuous?

  1. $f(x) = x^3 -4x + 1$
  2. $g(x) = \sqrt{x} + 2$
  3. $b(x) = \dfrac{x^2 + 2}{x^2 + 4}$
Click to reveal the answer. 1. $(\infty, \infty)$ since it's a polynomial 2. $[0, \infty)$ since this is the domain of the function. 3. $(\infty, \infty)$. Sneaky! Since the denominator is always greater than or equal to 4, it can never be 0.

Continuity and function algebra #

If $f$ and $g$ are continuous functions and the number $c$ is a constant, then the following combinations of functions are also continuous:

  1. $f + g$
  2. $f-g$
  3. $fg$
  4. $cf$
  5. $f/g$ if $g(x) \ne 0$.

Additionally, if $f$ is continuous at $b$ and $\displaystyle \lim_{x\to a} g(x) = b$, then $$ \lim_{x\to a} f(g(x)) = f\left( \lim_{x\to a} g(x) \right) = f(b) $$

… that is, if $f$ is continuous there, we can pass the limit inside the function.

Example #

Evaluate $\displaystyle \lim_{x\to 2} \sin\left( \frac\pi4 x \right)$

Solution: Since $\displaystyle\lim_{x\to 2} \frac{\pi}{4}x = \frac\pi2$ and $\sin(x)$ is continuous at $x=\frac\pi2$, we can pass the limit inside the continuous sine function: $$ \lim_{x\to 2} \sin\left( \frac\pi4 x \right) = \sin\left( \lim_{x\to 2} \frac\pi4 x \right) = \sin\left( \frac\pi2 \right) = 1 $$

Example: #

Evaluate $\displaystyle \lim_{x\to 0} x^4 \cos\left( \frac1x \right)$

For this one, $\frac1x$ is discontinuous at $x=0$, so we cannot move the limit inside the cosine function! We need other tools to evaluate the limit. Give it a try (hint: Squeeze Theorem).

Intermediate Value Theorem #

Suppose that $f$ is continuous on a closed interval $[a, b]$ and let $N$ be any number between $f(a)$ and $f(b)$. Then there exists a number $c$ in $(a, b)$ such that $f(c) = N$.

*somewhere between $a$ and $b$ there’s a number for which $f(c)$ hits the desired $y$-value of $N$.

Intuitively, if I’m Jabara Hall and want to walk to the Pod by the pond, I have to cross Perimeter2 road. My travel is continuous - I can’t teleport or use a jet pack.

An application example #

Show that the function $f(x) = x^4 + x- 3$ has a zero in the interval (1,2)

Now practice! #

Head over to WebAssign and work on section 1.8. If you have questions, let me know!

  1. In this class, we won’t be studying exponential and logarithmic functions until chapter 6 for… calculus reasons… They’re continuous on their domains, though, FYI. ↩︎

  2. I’ve been here over ten years now… at this point the road is called Midcampus Drive, but old habits die hard… now get off my lawn! ↩︎