Wemazi

Topic 1.3

Read a one-sided limit from a graph

Builds on recognizing when a limit does not exist. 6 minutes.

Approach 2 from one side

The curve breaks at \(x = 2\). The left part is a rising line that ends at an open ring at height 3. The right part is a falling line that starts at an open ring at height 4. An open ring marks a point that the curve does not include.

Press Approach 2 from the left. The orange point moves along the left part, so at \(x = 1.9\) its height is 2.9 and at \(x = 1.99\) it is 2.99. The heights come closer to 3.

Press Approach 2 from the right. At \(x = 2.1\) the height is 3.9, and at \(x = 2.01\) it is 3.99. The heights come closer to 4.

Each side approaches its own height, which is called a one-sided limit.

Any break

The curve breaks at an input \(c\). The left part heads toward a height \(\cy{L}\), and the right part heads toward a height \(\cy{R}\).

Use the slider Distance from c to move the orange point toward \(c\) along the left part. Its height comes closer to \(L\), although the curve never reaches the open ring. Then put the point on the right part, where its height comes closer to \(R\).

Move the Left height slider. Only the left part of the curve moves, so only the height \(L\) that it approaches changes. Each one-sided limit depends only on its own part of the curve.

Notation

  1. Write \(x \to c^-\) when \(x\) approaches \(c\) from the left, with \(x\) less than \(c\). The limit from the left is the height that the left part approaches.

    \[\lim_{x \to c^-} f(x) = \cy{L}\]

  2. Write \(x \to c^+\) when \(x\) approaches \(c\) from the right, with \(x\) greater than \(c\). The limit from the right is the height that the right part approaches.

    \[\lim_{x \to c^+} f(x) = \cy{R}\]

  3. A one-sided limit is the number that \(f(x)\) comes close to when \(x\) comes close to \(c\) on the named side only. It exists when that part of the curve heads toward one finite height.

    The value of \(f\) at \(c\) itself is not used.

When it fails

In this picture the right part climbs without bound as \(x\) comes close to \(c\) from the right. Its heights do not approach any number, so the limit from the right does not exist as a number.

The left part is unchanged, so the limit from the left still exists. The last skill of this topic reads the direction of such a climb.

Example 1: a break at 3

Let \(f(x) = 2x - 1\) for \(x < 3\) and \(f(x) = \tfrac{x}{2} + 1\) for \(x > 3\). Find \(\lim_{x \to 3^-} f(x)\) and \(\lim_{x \to 3^+} f(x)\).

  1. For the limit from the left, use only the part with \(x < 3\), where \(f(x) = 2x - 1\). At \(x = 2.9\) the height is 4.8, and at \(x = 2.99\) it is 4.98.

    \[f(2.9) = 4.8, \quad f(2.99) = 4.98\]

  2. The heights approach the open ring. The line has no break at 3, so the ring is at height \(2(3) - 1 = 5\).

    \[\lim_{x \to 3^-} f(x) = \cy{5}\]

  3. For the limit from the right, use only the part with \(x > 3\), where \(f(x) = \tfrac{x}{2} + 1\). At \(x = 3.01\) the height is 2.505.

  4. The heights approach the second open ring, at height \(\tfrac{3}{2} + 1 = \tfrac{5}{2}\).

    \[\lim_{x \to 3^+} f(x) = \cy{\tfrac{5}{2}}\]

Example 2: a break at 1

Let \(g(x) = x^2 - 2\) for \(x < 1\) and \(g(x) = 4 - 2x\) for \(x > 1\). Find \(\lim_{x \to 1^-} g(x)\) and \(\lim_{x \to 1^+} g(x)\).

  1. The limit from the left uses inputs less than 1, so only the left part, the parabola \(y = x^2 - 2\), counts.

  2. A tempting wrong move: Read the height of the part that starts at \(x = 1\): the limit from the left is 2.

    \[\lim_{x \to 1^-} g(x) = 2\]

    The orange point comes in from the left, so it stays on the parabola. The parabola ends near height \(-1\), far below the red line at 2. The ring at height 2 belongs to the part where \(x > 1\).

  3. Use the left part. At \(x = 0.99\) the height is \(-1.0199\), and the heights approach \(1^2 - 2 = -1\).

    \[\lim_{x \to 1^-} g(x) = \cy{-1}\]

  4. For the limit from the right, use the part with \(x > 1\). Its heights approach \(4 - 2(1) = 2\).

    \[\lim_{x \to 1^+} g(x) = \cy{2}\]

Practice

Let \(h(x) = 3 - x\) for \(x < -1\) and \(h(x) = x^2 - 3\) for \(x > -1\). Find \(\lim_{x \to -1^-} h(x)\) and \(\lim_{x \to -1^+} h(x)\).

Slide the orange point toward \(x = -1\) along one part at a time. Read the height that this part approaches.

Solution

  1. From the left, use \(h(x) = 3 - x\). Its heights approach \(3 - (-1) = 4\).

    \[\lim_{x \to -1^-} h(x) = \cy{4}\]

  2. From the right, use \(h(x) = x^2 - 3\). Its heights approach \((-1)^2 - 3 = -2\).

    \[\lim_{x \to -1^+} h(x) = \cy{-2}\]

Questions

Use the graph for the first two questions. A filled dot shows the value of f at that input.

  1. What is \(\lim_{x \to -2^-} f(x)\)?

    • \(3\)
    • \(-1\)
    • \(1\)
    Answer

    \(3\)

  2. What is \(\lim_{x \to 1^+} f(x)\)?

    • \(0\)
    • \(2\)
    • \(1\)
    Answer

    \(0\)

  3. Let \(h(x) = x^2 - 5\) for \(x < 2\) and \(h(x) = 3x - 2\) for \(x \geq 2\). What is \(\lim_{x \to 2^+} h(x)\)?

    • \(4\)
    • \(-1\)
    • The limit does not exist.
    Answer

    \(4\)

Summary

You follow the part of the curve on the named side of the input, and you read the height that this part approaches.

This works because a one-sided limit uses only the inputs on its side, not the other side or the value at the input.

Use it when a limit has a plus or minus sign, or when a graph breaks at the input.

This is one of four skills in topic 1.3 of Which & Why: AP Calculus.