Wemazi

Topic 1.13

Find the missing value

Builds on deciding whether a break can be filled. 7 minutes.

Heights that approach a number

The curve shows \(f(x) = \dfrac{x^3 - 8}{x - 2}\), which gives 0/0 at \(x = 2\), so the curve has a ring there. The question is which value fills the ring.

Move the two orange points toward 2. At \(x = 1.9\) and \(x = 2.1\) the heights are 11.41 and 12.61, and at \(x = 1.99\) and \(x = 2.01\) they are about 11.94 and 12.06.

The heights come close to 12. A table of heights only suggests a value, because a calculator rounds and cannot give the height at 2 itself.

The exact value comes from algebra, which removes the zero from the bottom so that the formula can be evaluated at the break.

Remove the zero

The top and the bottom are both 0 at \(c\), so both contain the factor \(x - c\). The letters \(p(x)\) and \(q(x)\) stand for what remains of the top and of the bottom.

Press the button. The shared factor cancels, and the new fraction \(g\) equals \(f\) at every input except \(c\).

If \(q(c)\) is not 0, then \(g\) has no break at \(c\), so its limit at \(c\) is \(g(c)\).

A limit ignores the point at \(c\) itself, so \(f\) has the same limit. The value \(g(c)\) is the missing value, and the orange dot shows it.

Notation

  1. Substitute \(c\) into the formula. If the top and the bottom are both 0, they share a factor \(x - c\).

  2. Rewrite the formula so that the factor \(x - c\) shows on the top and the bottom. Factoring does this for a polynomial. Multiplying by the conjugate does this for a square root.

    The conjugate of \(\sqrt{a} - b\) is \(\sqrt{a} + b\).

  3. Cancel the factor. This is allowed because a limit uses only the values with \(x \neq c\).

    \[f(x) = g(x) \quad (x \neq c)\]

  4. Substitute \(c\) into the simpler form \(g\). The result is the missing value.

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

  5. Some formulas contain a sine and cannot be rewritten with algebra. For these, use a known limit, with \(\theta\) in radians.

    \[\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = \cy{1}\]

When it fails

If a factor \(x - c\) stays in the bottom after the canceling, substituting \(c\) divides by 0, so there is no missing value.

Press the button. Beside the vertical line at \(c\), the curve rises on one side and falls on the other, so it approaches no single height.

Example 1: a square root

Find the value that fills the break of \(f(x) = \dfrac{\sqrt{x + 16} - 4}{x}\) at \(x = 0\).

  1. At \(x = 0\) the top is \(\sqrt{16} - 4 = 0\), and the bottom is 0. No factor \(x\) shows on the top, because the top contains a square root.

  2. Multiply the top and the bottom by the conjugate \(\sqrt{x + 16} + 4\). This multiplies the fraction by 1. Because \((a - b)(a + b) = a^2 - b^2\), the new top is \((x + 16) - 16 = x\).

    \[\begin{aligned} &\frac{(\sqrt{x+16}-4)(\sqrt{x+16}+4)}{x\,(\sqrt{x+16}+4)} \\ &\quad = \frac{x}{x\,(\sqrt{x+16}+4)} \end{aligned}\]

  3. Cancel \(x\), which is not 0 near 0. Then substitute 0 into what remains.

    \[\frac{1}{\sqrt{16}+4} = \cy{\tfrac18}\]

  4. As a check, the height at \(x = 0.01\) is about 0.125. The break is filled with \(\tfrac18\).

Example 2: a sine

Find the value that fills the break of \(f(x) = \dfrac{\sin 4x}{7x}\) at \(x = 0\).

  1. At \(x = 0\) the bottom is 0, and \(\sin 0\) is also 0. Algebra cannot remove this zero, so use the known limit for a sine.

  2. A tempting wrong move: Cancel the \(x\) in the top with the \(x\) in the bottom.

    \[\frac{\sin 4x}{7x} = \frac{\sin 4}{7}\]

    The \(x\) inside \(\sin 4x\) is not a factor of the top, so it cannot cancel. The result \(\tfrac{\sin 4}{7}\) is about \(-0.11\), below the axis, but the curve near 0 is above the axis, at height 0.56 when \(x = 0.1\).

  3. Write the bottom as \(7x = \tfrac74 \cdot 4x\), so that the same expression \(4x\) appears in the sine and below it.

    \[\frac{\sin 4x}{7x} = \frac{4}{7}\cdot\frac{\sin 4x}{4x}\]

  4. As \(x\) approaches 0, \(4x\) approaches 0, so \(\dfrac{\sin 4x}{4x}\) approaches 1. The missing value is \(\tfrac47\) times 1.

    \[\lim_{x\to 0} \frac{\sin 4x}{7x} = \cy{\tfrac47}\]

Practice

Find the value that fills the break of \(g(h) = \dfrac{\frac{1}{2+h} - \frac{1}{2}}{h}\) at \(h = 0\).

Move the points toward \(h = 0\) and read the height that they approach. Then combine the two fractions on the top, so that a factor \(h\) appears.

Solution

  1. Combine the fractions on the top, using the common denominator \(2(2 + h)\).

    \[\begin{aligned} \frac{1}{2+h} - \frac12 &= \frac{2 - (2+h)}{2(2+h)} \\ &= \frac{-h}{2(2+h)} \end{aligned}\]

  2. Divide by \(h\), and cancel \(h\), which is not 0 near 0.

    \[g(h) = \frac{-1}{2(2+h)}\]

  3. Substitute \(h = 0\). The missing value is \(-\tfrac14\).

    \[g(0) = \cy{-\tfrac14}\]

Questions

The graph shows \(w(x) = \dfrac{x^2 + 2x - 8}{x - 2}\). It has a ring at \(x = 2\).

  1. What height do the orange points approach as they move toward \(x = 2\)?

    • 6
    • 2
    • 0
    • 4
    Answer

    6

  2. Which expression equals \(w(x)\) for every \(x\) except 2?

    • \(x + 4\)
    • \(x - 4\)
    • \(x + 2\)
    Answer

    \(x + 4\)

  3. Find the value that fills the break of \(m(x) = \dfrac{x - 9}{\sqrt{x} - 3}\) at \(x = 9\).

    • 6
    • 0
    • 3
    • \(\tfrac16\)
    Answer

    6

Summary

You rewrite the formula so that the factor that makes the bottom 0 cancels, and you substitute the input of the break into the result.

The answer is correct because the original and the simpler formula agree near the break, so they have the same limit.

Algebra gives the exact value, and a table or a graph only confirms it.

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