One-sided limits and absolute values

Left of a, right of a

What is \displaystyle\lim_{x\to 2}\frac{|x-2|}{x-2}? To the right of 2 the expression equals 1. To the left of 2 it equals -1. A limit can have only one value, so what does that make this one?

Below, the indigo point approaches a from the left and the ochre point approaches from the right, always the same distance away. Move them closer. Do they head toward the same height?

Drag either point toward a, or use the controls. Both points stay the same distance from a.

Limit
Distance
f(a − distance)
f(a + distance)
Two-sided limit

Values of f(x) near a, from both sides
Distance from af(a − distance)f(a + distance)

One-sided limits

The notation \lim_{x\to a^-}f(x) means the limit as x approaches a using only values x. Likewise \lim_{x\to a^+}f(x) uses only x>a. The two-sided limit uses both, so it can exist only when the two sides agree:

\lim_{x\to a}f(x)=L\quad\Longleftrightarrow\quad \lim_{x\to a^-}f(x)=L\ \text{ and }\ \lim_{x\to a^+}f(x)=L

As always, the value f(a) plays no part. Choose the piecewise function with a jump in the graph above. What is f(2), and does it match either side?

Absolute values are two cases

By definition, |u|=u when u\ge 0 and |u|=-u when u<0. So an absolute value hides a piecewise function, and the place where the inside equals 0 is where the two cases meet. If that place is a, each side of a needs its own formula.

Take \lim_{x\to 2}\frac{|x^2-4|}{x-2}. Just to the right of 2, x^2-4>0, so

\lim_{x\to 2^+}\frac{x^2-4}{x-2}=\lim_{x\to 2^+}(x+2)=4.

Just to the left of 2, x^2-4<0, so |x^2-4|=-(x^2-4) and

\lim_{x\to 2^-}\frac{-(x^2-4)}{x-2}=\lim_{x\to 2^-}-(x+2)=-4.

The one-sided limits are 4 and -4. They differ, so the two-sided limit does not exist.

Solve it step by step

Work out each side separately, then decide whether the two-sided limit exists. Type a number, or type inf or -inf when a side grows without bound. The score counts steps you answer correctly on the first try.

Problems solved0
First-try steps0 of 0

    Common mistakes

    Dropping the bars

    |x-2| is not x-2 and not x+2. It equals x-2 on one side of 2 and -(x-2) on the other.

    Checking one side only

    A table that approaches from the right alone can make a limit look like it exists. Always check both sides.

    Using f(a)

    A piecewise function may define f(a) with one of its pieces. That value says nothing about the limit.

    Treating ∞ as a number

    Writing \lim_{x\to a^+}f(x)=\infty describes how the limit fails to exist. It is not a value the limit equals.