Continuity and Piecewise Functions

Joining a value to its neighbors

Continuity at $a$ means the value of a function agrees with the trend of nearby values. There are three checks: $f(a)$ must be defined, the finite limit at $a$ must exist, and that limit must equal $f(a)$ . Missing any one of these breaks continuity.

Polynomials are continuous everywhere. Rational functions are continuous wherever their denominator is nonzero. Roots and logarithms are continuous on their real domains. Compositions preserve continuity when the inner output belongs to the outer function's domain. At an endpoint of a domain, use the appropriate one-sided limit.

Classify the obstruction

A removable discontinuity has a finite common limit but the point is missing or has the wrong value. A jump has distinct finite one-sided limits. An infinite discontinuity involves unbounded values near the point. These descriptions identify what, if anything, changing a single value can repair.

Changing $f(a)$ can fix a removable hole. It cannot reconcile distinct one-sided limits, and it cannot turn unbounded nearby values into a finite continuous value.

Worked example: join two formulas

Suppose $f(x)=x^2$ for $x<2$ and $f(x)=mx+1$ for $x\ge2$ . The left-hand limit is $4$ . The right-hand limit and actual value are both $2m+1$ . Continuity requires $2m+1=4$ , so $m=3/2$ .

Notice the order: identify which expression defines the point, calculate the two nearby trends, then equate them. Substituting into the wrong branch can give a plausible answer to a different problem.

Worked example: continuous does not mean smooth

The function $|x|$ is continuous at zero: both sides approach zero and $|0|=0$ . Its left slope is $-1$ and its right slope is $1$ , however, so it is not differentiable there. Every differentiable function is continuous at that point, but the reverse implication fails.

For the rational expression $(x^2-1)/(x-1)$ , the original domain excludes $1$ . Nearby it equals $x+1$ , so defining a new value of $2$ at $1$ gives a continuous extension. The original formula and its extension have different domains.

Using continuity responsibly

Continuity allows substitution in limits and supports existence theorems. It does not imply boundedness on an open interval: $1/x$ is continuous on $(0,1)$ but unbounded there. A continuous function on a closed bounded interval is bounded and attains its maximum and minimum.

Practice

Set $f(x)=2x+k$ for $x<1$ and $f(x)=x^2$ for $x\ge1$ . Find $k$ for continuity.
Can redefining $1/x$ at zero make it continuous there?
Give a continuous function with a corner and explain what fails at that corner.

Show worked solutions

The left limit is $2+k$ , while the value and right limit are $1$ . Thus $k=-1$ .
No. There is no finite common limit at zero, regardless of a chosen point value.
$|x|$ is continuous at zero, but its unequal one-sided derivatives prevent differentiability.

Explore

Value versus limit

Try this. Compare a hole and a jump. In the defined-value mode, move f(0) to 0: continuity requires the value to equal the common limit.

BehaviorDefined value f(0)0

Both sides approach 0, but f(0) is undefined. The hole is removable.