Wemazi

Topic 1.6

Use a conjugate with radicals

Builds on factoring and canceling to reveal a limit. 6 minutes.

A hole that a square root hides

The chalk curve shows \(\dfrac{\sqrt{x+11}-4}{x-5}\). At \(x=5\) the top is \(\sqrt{16}-4=0\) and the bottom is 0, so the curve has a hole there.

The top has a square root, so you cannot factor it as you factor a polynomial. Drag the orange dot toward 5. At x = 4.9 the height is about 0.1252, and at x = 5.1 it is about 0.1248, so the heights approach 0.125.

Press the button to draw the orange dashed curve \(\dfrac{1}{\sqrt{x+11}+4}\). It has no square root in the top, and it has the same heights as the chalk curve at every input except 5.

The conjugate removes the square root

The chalk curve is now a general quotient whose top is \(\sqrt{u}-b\), where \(u\) is an expression in \(x\). The top and the bottom are both 0 at the input \(c\), so the curve has a hole there.

The conjugate of \(\sqrt{u}-b\) is \(\sqrt{u}+b\), which has the same two terms with the opposite sign between them. The product of the two is \(u-b^2\), and it has no square root.

Press the first button. This multiplies the top and the bottom by the conjugate, which is the same as multiplying the quotient by 1. The dashed curve lies on the chalk curve and keeps the hole.

The new top, \(u-b^2\), is 0 at \(c\), so it has the factor \(x-c\). Press the second button to cancel that factor. The dashed curve now has no hole, so its limit is its value at \(c\).

Notation

  1. Substitute \(c\). The top has a square root, and the form is 0 over 0.

  2. Write the conjugate of the radical expression. The product of the two is a difference of squares, so the square root disappears.

    \[\bigl(\sqrt{\cq{u}}-b\bigr)\bigl(\sqrt{\cq{u}}+b\bigr)=\cq{u}-b^2\]

  3. Multiply the top and the bottom by the conjugate. This multiplies the quotient by 1, so the values do not change.

    \[\begin{gathered}\frac{\sqrt{u}-b}{x-c}\cdot\frac{\sqrt{u}+b}{\sqrt{u}+b}\\ =\frac{u-b^2}{(x-c)\bigl(\sqrt{u}+b\bigr)}\end{gathered}\]

  4. The new top has the form \(k(x-c)\). Cancel \(x-c\) for every \(x\) different from \(c\). The new bottom is not 0 at \(c\), so substitute.

    \[\frac{\cy{k}(x-c)}{(x-c)\bigl(\sqrt{u}+b\bigr)}=\frac{\cy{k}}{\sqrt{u}+b}\]

When it fails

If the multiplier has the same sign as the top, the new top is \((\sqrt{x+11}-4)^2\), which expands to \(x+27-8\sqrt{x+11}\), so the square root stays.

The new top and the new bottom are both 0 at x = 5. The top has no factor x − 5 to cancel, so the dashed curve keeps the hole. Only the conjugate, with the opposite sign, removes the square root.

Example 1: A square root in the bottom

Find the limit of \(\dfrac{x-16}{\sqrt{x}-4}\) as \(x\) approaches 16.

  1. Substitute 16. The top and the bottom are both 0.

    \[16-16=0,\qquad \sqrt{16}-4=0\]

  2. The square root is in the bottom. Multiply the top and the bottom by the conjugate \(\sqrt{x}+4\).

    \[\frac{(x-16)(\sqrt{x}+4)}{(\sqrt{x}-4)(\sqrt{x}+4)}\]

  3. The bottom is a difference of squares, \((\sqrt{x})^2-16\), which equals \(x-16\).

    \[\frac{(x-16)(\sqrt{x}+4)}{x-16}\]

  4. Cancel \(x-16\) for every \(x\) different from 16. Then substitute.

    \[\sqrt{x}+4\ \to\ \sqrt{16}+4=8\]

Example 2: Multiply the whole quotient

Find the limit of \(\dfrac{\sqrt{3x+3}-3}{x-2}\) as \(x\) approaches 2.

  1. Substitute 2. The top is \(\sqrt{9}-3=0\), and the bottom is 0.

    \[\sqrt{3(2)+3}-3=0\]

  2. A tempting wrong move: Multiply only the top by the conjugate \(\sqrt{3x+3}+3\). Then cancel.

    \[\begin{gathered}\frac{(3x+3)-9}{x-2}\\ =\frac{3(x-2)}{x-2}=3\end{gathered}\]

    Multiplying the top alone changes the quotient. The new quotient is the old one times √(3x + 3) + 3, which is close to 6 near x = 2. The red dot at height 3 is 6 times the height of the open ring.

  3. Multiply the top and the bottom by the conjugate. Cancel \(x-2\). Then substitute.

    \[\begin{gathered}\frac{3(x-2)}{(x-2)\bigl(\sqrt{3x+3}+3\bigr)}\\ =\frac{3}{\sqrt{3x+3}+3}\ \to\ \frac{3}{6}=\frac12\end{gathered}\]

Practice

Find the limit of \(\dfrac{2-\sqrt{x+1}}{x-3}\) as \(x\) approaches 3.

The top is 2 − √(x + 1), so its conjugate is 2 + √(x + 1). Multiply the top and the bottom by it.

Solution

  1. The new top is \(4-(x+1)=3-x\), which is \(-(x-3)\). Cancel \(x-3\).

    \[\begin{gathered}\frac{3-x}{(x-3)\bigl(2+\sqrt{x+1}\bigr)}\\ =\frac{-1}{2+\sqrt{x+1}}\end{gathered}\]

  2. The new bottom is not 0 at 3, so substitute.

    \[\frac{-1}{2+2}=-\frac14\]

Questions

The chalk curve is \(g(x)=\dfrac{\sqrt{x+3}-2}{x-1}\). The orange dashed curve is what remains after you multiply by the conjugate and cancel. The open ring marks a missing point of \(g\).

  1. What is the limit of \(g(x)\) as \(x\) approaches 1?

    • \(\tfrac14\)
    • 0
    • \(\tfrac12\)
    Answer

    \(\tfrac14\)

  2. At which input do the chalk curve and the dashed curve differ?

    • x = 1
    • x = 3
    • x = 0
    Answer

    x = 1

  3. What is the limit of \(\dfrac{\sqrt{x+2}-3}{x-7}\) as \(x\) approaches 7?

    • \(\tfrac16\)
    • 1
    • \(\tfrac13\)
    • 0
    Answer

    \(\tfrac16\)

Summary

You multiply the top and the bottom by the conjugate of the square root expression, cancel the common factor, and substitute.

The conjugate removes the square root, and because the conjugate over itself equals 1, the heights do not change away from the hole.

Use it when substitution gives 0 over 0 and a square root hides the factor that you need.

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