Wemazi

Topic 1.2

Separate the limit from the function value

Builds on comparing the two sides of a limit. 6 minutes.

A point that can move

The graph shows \(f(x) = \dfrac{x^2 + x - 6}{x - 2}\). At \(x = 2\) the formula gives 0 divided by 0, so \(f(2)\) has no value. For every other input, the top factors as \((x + 3)(x - 2)\), so \(f(x) = x + 3\). The graph is the line \(y = x + 3\) with an open ring at the missing point.

Drag the left orange dot toward 2. The right dot moves toward 2 from the other side, and both outputs approach 5, so the limit is 5.

The buttons set \(f(2)\) to 5 or 1, shown by a chalk dot, or remove it. The curve near 2 does not change, so the limit is 5 in all three cases.

The limit and the value

This curve has no numbers. Near an input \(c\), it approaches a height \(L\), the limit, which the orange line marks.

The chalk dot shows the value \(f(c)\), which can be missing, equal to \(L\), or different from \(L\). Move it with the slider.

The orange line stays where it is in every case, because the limit depends only on the curve near \(c\), not on the dot.

Notation

  1. The limit \(L\) is the number that the outputs near \(c\) approach, which the orange line marks.

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

  2. The value \(f(c)\) is the output at \(c\) itself, which the chalk dot shows. It can be a number, or it can be missing.

    \[f(c)\]

  3. The limit does not use \(f(c)\), so the two numbers are separate. When the dot moves, the orange line stays where it is.

  4. The two numbers are equal only when the dot fills the ring.

When it fails

Move the dot off the curve. The orange line stays at \(L\), but the value is now a different number.

When the limit and the value differ, the graph has a break at c.

Example 1: a missing value

Find the limit of \(f(x) = \dfrac{x^2 - 25}{x - 5}\) as \(x\) approaches 5. Then state \(f(5)\).

  1. At \(x = 5\) the formula gives 0 divided by 0, so \(f(5)\) has no value.

  2. For \(x \neq 5\), factor the top. Then cancel \(x - 5\), which is allowed because \(x - 5\) is not 0. The faint dashed line shows the simpler formula.

    \[f(x) = \dfrac{(x - 5)(x + 5)}{x - 5} = x + 5\]

  3. Near 5 the outputs approach 10, so the limit is 10, even though \(f(5)\) is not defined.

    \[\lim_{x \to 5} f(x) = \cy{10}\]

Example 2: a different value

A function has \(g(x) = x^2\) for \(x \neq 3\) and \(g(3) = 1\). Find \(\lim_{x \to 3} g(x)\).

  1. The value \(g(3) = 1\) is the chalk dot at \((3, 1)\).

  2. A tempting wrong move: The value is 1, so the limit is 1.

    \[\lim_{x \to 3} g(x) = 1\]

    The red line at height 1 passes through the chalk dot, far below the curve. The outputs near 3 are near 9, because \(g(2.99) = 8.9401\) and \(g(3.01) = 9.0601\).

  3. The outputs near 3 approach 9, so the limit is 9. The value \(g(3) = 1\) is a different number.

    \[\lim_{x \to 3} g(x) = \cy{9}\]

Practice

Let \(k(x) = \dfrac{x^2 - 9x + 20}{x - 4}\) for \(x \neq 4\), and let \(k(4) = 3\). Find \(\lim_{x \to 4} k(x)\) and compare it with \(k(4)\).

Factor the top so that you can cancel \(x - 4\). The limit is the height of the ring, not the height of the chalk dot.

Solution

  1. For \(x \neq 4\), factor the top and cancel \(x - 4\).

    \[k(x) = \dfrac{(x - 4)(x - 5)}{x - 4} = x - 5\]

  2. As \(x\) approaches 4, the outputs \(x - 5\) approach \(-1\), so the limit is \(-1\).

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

  3. The value \(k(4) = 3\) is the chalk dot above the ring, so the limit and the value are different numbers.

Questions

The graph shows a function \(r\) that follows the line \(y = \tfrac12 x + 1\), with a ring at \(x = 1\) and a ring at \(x = 3\). A chalk dot at \(x = 3\) shows \(r(3)\).

  1. At \(x = 3\), what are the limit and the value \(r(3)\)?

    • Limit 2.5, value 4
    • Limit 4, value 4
    • Limit 2.5, value 2.5
    Answer

    Limit 2.5, value 4

  2. Which statement is true at \(x = 1\)?

    • The limit is 1.5, but \(r(1)\) is not defined
    • The limit does not exist, because \(r(1)\) is not defined
    • The limit is 1
    Answer

    The limit is 1.5, but \(r(1)\) is not defined

  3. Find \(\lim_{x \to 3} \dfrac{x^2 - 6x + 9}{x - 3}\).

    • 0
    • It does not exist, because it is \(0 \div 0\)
    • 3
    Answer

    0

Summary

You find a limit from the outputs near an input, and you read the value at that input separately.

This works because the limit uses only the inputs near a point, never the point itself, so the value there can be missing or different.

This matters when a formula gives 0 ÷ 0, or when a function has a different value at one point.

This is one of four skills in topic 1.2 of Which & Why: AP Calculus.