Continuity and removable discontinuities

Understand the idea

Continuity at an interior point requires a defined function value, an existing finite limit, and equality of that limit to the function value. At an included endpoint use the relevant one-sided limit. A removable discontinuity can be fixed by assigning the limiting value. Jumps and infinite discontinuities cannot be fixed by changing a single function value. For piecewise rules compare the appropriate formulas at every join.

f\text{ continuous at }a\iff f(a)\text{ exists and }\lim_{x\to a}f(x)=f(a)

Worked example

Define $\displaystyle f(x)=(x^2-4)/(x-2)$ for $\displaystyle x\ne2$ and $\displaystyle f(2)=k$ . Find $\displaystyle k$ for continuity.

For

\displaystyle x\ne2

, the expression equals

\displaystyle x+2

. Its limit at

\displaystyle 2

is

\displaystyle 4

, so continuity requires

\displaystyle k=4

. This fills the hole and creates a continuous extension.

Try it yourself

For $\displaystyle f(x)=x+1$ when $\displaystyle x<1$ and $\displaystyle f(x)=kx$ when $\displaystyle x\ge1$ , find $\displaystyle k$ for continuity.

Show solution and reasoning

The left-hand limit is $\displaystyle 2$ . The right-hand limit and $\displaystyle f(1)$ are both $\displaystyle k$ . Thus $\displaystyle k=2$ .