Why Continuity?

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We would like to study nice well-behaved functions; functions that are smooth and that don’t do strange things. Yeah, well good luck with that.

One of the things that might be nice is that you could draw the graph of a function from one end of its domain to the other without taking your pen off the paper. And a lot of functions are like that, but not all.

Some functions have holes in them, others jump from one y-value to another without hitting points in between. Some “go off to infinity” and come right back; others go off the top of the graph and come back from the bottom. Some go really crazy around a point.  

Functions that you can draw from one end of their domain to the other without lifting your pen are said to be continuous.

More mathematically: A function is continuous at a point (that is, at a single value of x) if, and only if, as you travel along the graph towards the value (from either side of the x-value), the y­-values on the graph are approaching the ­y­-value at the point. In symbols:  \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right).

A function is continuous on an interval if, and only if, it is continuous at every value in the interval.

Wait! What?? I have to check all the points??

Technically, yes; practically, no. Most of the time you can easily show that a function is continuous everywhere by looking at its limit in general.

Moreover, you will learn to see where a function is not continuous. This is an important skill: looking at a function and suspecting there is a problem with continuity.

Take a quick look at some of the problems that functions may have at a point. Graph these on your calculator. They all have a “problem” at x = 3. Graph each example and you will see what they look like. Try to figure out why they have a “problem” and what causes it.  

  •  \displaystyle f\left( x \right)=\frac{1}{{x-3}}.
  •  \displaystyle g\left( x \right)=\frac{1}{{{{{\left( {x-3} \right)}}^{2}}}}
  • \displaystyle h\left( x \right)=\frac{{{{x}^{2}}-3x}}{{x-3}}. This function has a single hole in the graph at (3, 3); It one may be difficult to see. Try using ZDecimal. A single point is missing because there is no value at x= 3 because the denominator is zero.
  •  \displaystyle j\left( x \right)=\frac{{{{x}^{2}}\sqrt{{{{x}^{2}}-6x+9}}}}{{2x-6}}.
  • \displaystyle k\left( x \right)=\cos \left( {\frac{1}{{x-3}}} \right) Zoom in several times at (3, 0) where the function has no value.
  • \displaystyle m\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}} & {x<3} \\ {4-x} & {x\ge 3} \end{array}} \right.

Learn to suspect that a function may have a discontinuity. (It’s not always at x = 3) The problem is often a zero denominator.

This is not just a game or some curious functions. One of the main tools of calculus called the derivative, which you will study next, is defined as the limit of a special function which is never continuous at the point you are interested in.

So, let’s continue on to continuity.

AP Calculus Course and Exam Description

Unit 1 topics 1-10 – 1.16, Unit 2 topic 2.4

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