Wemazi

Topic 1.6

Simplify a complex quotient

Builds on using a conjugate with radicals. 7 minutes.

The slope of a secant line to 1/(x + 1)

The curve shows \(y=\dfrac{1}{x+1}\). The first orange dot is fixed at \((3,\tfrac14)\), and the second orange dot is at the input \(x\). The orange line through both dots is a secant line.

The slope of the secant line is the change in height over the change in input, \(\dfrac{\frac{1}{x+1}-\frac14}{x-3}\), so it is a fraction inside a fraction. Drag the second dot toward the first one.

At x = 3.1 the slope is about −0.0610, and at x = 2.9 it is about −0.0641. The slopes approach −0.0625, which is the slope of the tangent line that the button shows.

At x = 3 the top and the bottom of the slope are both 0, so substitution fails. The next scenes rewrite the slope so that its limit shows.

A fraction inside a fraction

The first dot is fixed at the input \(\cq{c}\), and you drag the second dot at the input \(\cy{x}\). The slope of the secant line is the change in height divided by the change in input, \(x-c\).

If the curve is a fraction, the change in height is a difference of two fractions. At \(x=c\) the two fractions are equal, so the top and the bottom of the slope are both 0.

Write the two fractions in the top over one common denominator. The combined top is 0 at \(c\), so its numerator has the factor \(x-c\).

Dividing by \(x-c\) multiplies the bottom of the combined fraction by \(x-c\). Then \(x-c\) is a factor of both the numerator and the denominator, so it cancels for \(x\) different from \(c\). The canceled expression is defined at \(c\), so you find its limit by substituting \(c\).

Notation

  1. The slope of the secant line is \(\dfrac{\frac{1}{x+1}-\frac14}{x-3}\). Write the two fractions in its top over the common denominator \(4(x+1)\).

    \[\begin{aligned}\frac{1}{x+1}-\frac14 &=\frac{4-(x+1)}{4(x+1)}\\ &=\frac{3-x}{4(x+1)}\end{aligned}\]

  2. Dividing by \(x-3\) multiplies the bottom of the fraction by \(x-3\).

    \[\frac{\dfrac{3-x}{4(x+1)}}{x-3}=\frac{3-x}{4(x+1)(x-3)}\]

  3. Because \(3-x=-(x-3)\), you can cancel \(x-3\) for every \(x\) different from 3.

    \[\frac{-(x-3)}{4(x+1)(x-3)}=\frac{-1}{4(x+1)}\]

  4. The new expression is defined at 3, so substitute.

    \[\lim_{x\to 3}\frac{-1}{4(x+1)}=\cy{-\frac{1}{16}}\]

When it fails

The two dots are now the same point. A line needs two different points, so there is no secant line, and the slope has the form 0 over 0.

A limit uses inputs that approach 3 but are not 3. Rewrite the slope first. Then let \(x\) approach 3.

Example 1: A difference quotient

Find the limit of \(\dfrac{\frac{1}{6+h}-\frac16}{h}\) as \(h\) approaches 0.

  1. Write the top over the common denominator 6(6 + h).

    \[\begin{aligned}\frac{1}{6+h}-\frac16 &=\frac{6-(6+h)}{6(6+h)}\\ &=\frac{-h}{6(6+h)}\end{aligned}\]

  2. Dividing by \(h\) multiplies the bottom by \(h\). Cancel \(h\), because \(h\) is not 0 in a limit.

    \[\frac{-h}{6(6+h)\,h}=\frac{-1}{6(6+h)}\]

  3. Substitute \(h=0\).

    \[\frac{-1}{6\cdot 6}=-\frac{1}{36}\]

Example 2: Subtract fractions with a common denominator

Find the limit of \(\dfrac{\frac{3}{x-2}-\frac32}{x-4}\) as \(x\) approaches 4.

  1. Substitute 4. Both fractions in the top equal 3/2, so the top is 0, and the bottom 4 − 4 is 0 too.

    \[\frac32-\frac32=0\]

  2. A tempting wrong move: Subtract the tops and subtract the bottoms of the two fractions.

    \[\begin{aligned}\frac{3}{x-2}-\frac{3}{2} &=\frac{3-3}{(x-2)-2}\\ &=0\end{aligned}\]

    Fractions do not subtract this way. A top of 0 would make every secant line flat, like the red horizontal line. At x = 4.1 the slope is (3/2.1 − 3/2) divided by 0.1, which is about −0.714, so the secant line is not flat.

  3. Use the common denominator \(2(x-2)\), so the combined top has the factor \(x-4\). Divide by \(x-4\) and cancel it. Then substitute.

    \[\begin{gathered}\frac{3}{x-2}-\frac32=\frac{6-3(x-2)}{2(x-2)}\\ =\frac{-3(x-4)}{2(x-2)}\\ \frac{-3}{2(x-2)}\ \to\ -\frac34\end{gathered}\]

Practice

Find the limit of \(\dfrac{\frac1x+\frac15}{x+5}\) as \(x\) approaches \(-5\).

Write the top over the common denominator 5x. The new top has the factor x + 5.

Solution

  1. The top is \(\dfrac{5+x}{5x}\). Dividing by \(x+5\) multiplies the bottom by \(x+5\), and \(x+5\) cancels.

    \[\frac{\dfrac{x+5}{5x}}{x+5}=\frac{1}{5x}\]

  2. Substitute −5.

    \[\frac{1}{5(-5)}=-\frac{1}{25}\]

Questions

The graph shows \(y=\tfrac1x\). The first dot is fixed at \(x=2\). The orange line is the secant line from the first dot to the point at \(x=4\). The faint line is the tangent line at \(x=2\).

  1. What is the slope of the orange secant line?

    • \(-\tfrac18\)
    • \(\tfrac18\)
    • \(-\tfrac14\)
    Answer

    \(-\tfrac18\)

  2. As the second dot moves toward the first, the secant slopes approach the slope of the tangent line. What is the slope of the faint tangent line?

    • \(-\tfrac14\)
    • \(-\tfrac18\)
    • \(-\tfrac12\)
    Answer

    \(-\tfrac14\)

  3. What is the limit of \(\dfrac{\frac{2}{x+2}-\frac23}{x-1}\) as \(x\) approaches 1?

    • \(-\tfrac29\)
    • 0
    • \(\tfrac29\)
    • \(-\tfrac23\)
    Answer

    \(-\tfrac29\)

Summary

You write the fractions in the top over a common denominator, cancel the factor that makes the quotient 0 over 0, and substitute.

This works because the combined top has the factor in the bottom, and the canceled expression equals the quotient everywhere except at that point.

Use it for difference quotients of reciprocals, where the slope of a secant line is a fraction inside a fraction.

This is one of four skills in topic 1.6 of Which & Why: AP Calculus.