Section 2.5 · Continuity & IVT

Overview

Continuity means that there are “no breaks” in the graph. That is, if \(f(x)\) is continuous, you can draw its graph without picking up your pencil. Mathematically, being continuous at a value \(x=a\) means that \( \lim_{x\to a} f(x)=f(a) \).

Some of the main tasks you will do are: determine where a graph is continuous, explain why a function is discontinuous at a point, choose a parameter to make a piecewise function continuous, decide if a discontinuity is removable and patch it, and apply the IVT to guarantee a root.

Discontinuity at a point

Mark every x where there’s a hole (removable), jump, or vertical asymptote (infinite). The function is continuous on open intervals between those points.

Tool:
Your marks — Holes: ∅; Jumps: ∅; Asymptotes: ∅

Example: Piecewise function
Consider the function \( f(x)= \begin{cases} e^{x}+1,& x<0\\[2pt] x^2,& x\ge 0 \end{cases} \)
  1. \( \lim_{x\to 0^-} f(x)=e^{0}+1=2\).
  2. \( \lim_{x\to 0^+} f(x)=0^2=0\).
  3. Since the left hand limit and right hand limit are not equal, \(\lim_{x\to 0} f(x)\) does not exist. This means \(f\) is discontinuous at \(0\).

Making a function continuous

You may be asked to find a value of a parameter that makes a piecewise function continuous. Set the left-hand limit at the junction equal to the right-hand limit, then solve for the parameter.

You might also encounter functions that have removable discontinuities, and you'll be asked to find values that make such a function continuous.

Example
Consider the function \( f(x)= \begin{cases} c\,x^2-1, & x<2\\[2pt] 3x-5, & x\ge 2 \end{cases} \)

Continuity at \(x=2\) requires \( \lim_{x\to2^-} f(x) = \lim_{x\to2^+} f(x)=f(2)\). From the left, the limit is: \(4c-1\). From the right, the limit is: \(3(2)-5=1\), which is also the value of \(f(2)\). Therefore, we solve \(4c-1=1\), which implies that \(c=\tfrac12\).

Example
Consider the function \(\displaystyle f(x)=\frac{x^2-4}{x-2}\) for \(x\ne 2\).

This simplifies to \(x+2\), but since \(f(x)\) is undefined at \(x=2\), there is a hole. We can define the function

\(g(x)=\begin{cases} \dfrac{x^2-4}{x-2}, & x\ne 2\\ 4,& x=2\end{cases}\)

to fill the hole (since \(\lim_{x\to2}f(x)=4\)). Now the function \(g(x)\) is continuous at \(2\).

Intermediate Value Theorem (IVT)

IVT: —
Endpoints: \(a=\) —, \(b=\) —    \(f(a)=\) —, \(f(b)=\) —    \(L=\) —

IVT says: If \(f\) is continuous on the whole closed interval \([a,b]\) and \(L\) lies between \(f(a)\) and \(f(b)\), then there exists some \(c\in(a,b)\) with \(f(c)=L\). If there is a hole, a jump, or an asymptote, this guarantee can fail, since \(f\) is not continuous on \([a,b]\).

Example

Let \(h(x)=x^3-5x+1\) on \([0,1]\). Then \(h(0)=1>0\) and \(h(1)=-3<0\). By IVT there is \(c\) in \((0,1)\) with \(h(c)=0\). (We’re not asked to find \(c\), only to prove it exists.)