Why can no value repair the break at x = -2?

The curve falls without bound on the left and rises without bound on the right

The function is not defined at x = -2

The two sides end at different finite heights

Recognize a break that one value cannot repair

Builds on defining a continuous extension. 7 minutes.

The graph has a break at \(x = 2\). The left piece \(y = x^2 - 1\) ends at height 3, and the right piece \(y = 7 - x\) starts at height 5.

Move the orange dot up and down the line \(x = 2\). Each red segment shows the distance from the dot to one piece.

At height 3, the dot meets the left piece but is 2 below the right piece. At height 5, it meets the right piece but is 2 above the left one.

Because the pieces end at different heights, the two gaps are never 0 together, so no value at 2 joins them.

Press each button in turn, and try to place the orange dot so that it joins the graph.

In a jump, the two sides approach different heights, so the dot can meet one side at most.

In the second case, each side of the curve rises or falls without bound. Neither side approaches a height, so the dot has nothing to meet.

In an oscillation, the curve swings up and down faster and faster as it nears \(c\), so it approaches no single height.

In all three cases the two-sided limit, the one height that both sides approach, does not exist, so no single value joins the graph.

A jump discontinuity has two one-sided limits that are real numbers and are different.
\[\begin{aligned} \cy{L_1} &= \lim_{x\to c^-} f(x) \\ \cy{L_2} &= \lim_{x\to c^+} f(x) \\ L_1 &\neq L_2 \end{aligned}\]
An infinite discontinuity has at least one one-sided limit equal to \(\infty\) or \(-\infty\). The vertical line \(x = c\) is then a vertical asymptote.
\[\lim_{x\to c^+} f(x) = \pm\infty\]
The symbol \(\infty\) is not a number. It describes values that grow without bound, so such a limit does not exist as a real number.
In each case the two-sided limit does not exist, so no value of \(f(c)\) makes \(f\) continuous at \(c\). These breaks are non-removable.

A value at \(c\) does not repair a break, because a function can have a value at \(c\) and still be discontinuous there.

In this picture the dot meets the left side, but the right side still starts higher, so the graph stays broken.

Let \(w(x) = \dfrac{1}{x - 2}\) for \(x < 2\), and \(w(x) = 5\) for \(x \ge 2\). Can any change to the value at \(x = 2\) repair the break?

From the right, \(w(x) = 5\), so the limit is 5. The dot at \(w(2) = 5\) meets this side.
\[\lim_{x\to 2^+} w(x) = \cy{5}\]
From the left, \(x - 2\) is a small negative number, so \(\dfrac{1}{x - 2}\) is a large negative number. At \(x = 1.99\) the height is \(-100\), far below the figure.
\[\lim_{x\to 2^-} w(x) = \cy{-\infty}\]
The left side approaches no height, so no value at 2 can meet it, even though the dot meets the right side. The break is an infinite discontinuity.

Let \(g(x) = \dfrac{3}{(x - 4)^2}\) for \(x \neq 4\), and \(g(4) = 0\). Can any value at \(x = 4\) repair the break?

Near 4, the bottom \((x - 4)^2\) is a small positive number on both sides, so \(g(x)\) is a large positive number on both sides.
A tempting wrong move: Both sides rise without bound, so the two sides agree. Fill the break with their common value.
\[\lim_{x\to 4} g(x) = \infty \text{ is a value}\]
The height is 300 at \(x = 4.1\) and 30,000 at \(x = 4.01\). A dot at any real height, such as 6, lies far below the curve close to 4, and the red gap grows.
Two sides that rise without bound approach no real number. No value joins them, so the break is an infinite discontinuity.

Let \(h(x) = \dfrac{x^2 - 9}{|x - 3|}\) for \(x \neq 3\). Can any value at \(x = 3\) repair the break?

Move the points toward \(x = 3\) from each side. The absolute value changes the formula on one side, so find the limit of each side separately.

Solution

For \(x > 3\), \(|x - 3| = x - 3\). The top factors as \((x - 3)(x + 3)\), so \(h(x) = x + 3\), and the right side approaches 6.
\[\lim_{x\to 3^+} h(x) = \cy{6}\]
For \(x < 3\), \(|x - 3| = -(x - 3)\), so \(h(x) = -(x + 3)\), and the left side approaches \(-6\).
\[\lim_{x\to 3^-} h(x) = \cy{-6}\]
The two limits are different, so the break is a jump. No value repairs it.

The graph shows \(f\). The breaks are at \(x = -2\), \(x = 1\) and \(x = 4\).

You find the limit from each side and check whether the two limits are the same real number.

If the limits differ, or one side is unbounded, the two-sided limit does not exist, so no single value joins the graph.

The test also applies when the function has a value at the break, because that value does not decide the question.

This is one of four skills in topic 1.13 of Which & Why: AP Calculus.