7. Computing Limits

b. When Limit Laws Don't Apply

Limits without Laws

1. Limit Tricks

Divide by the Largest Term in the Denominator

Put Terms over a Common Denominator

Multiply by the Conjugate

e. Multiply by the Conjugate

The conjugate of a difference \(a-b\) is the sum \(a+b\). This is useful because their product is the difference of squares, \((a-b)(a+b)=a^2-b^2\). Consequently, if \(a\) and/or \(b\) is a square root, then \(a^2-b^2\) eliminates the square root.

This trick says if numerator or denominator contains a difference where one or both terms is a square root, then multiply the numerator and denominator by the conjugate. Multiply out the numerator or denominator which had the original difference. It is most appropriate for the indeterminate forms \(\dfrac{0}{0}\), \(\dfrac{\infty}{\infty}\), \(0\cdot\infty\) or \(\infty-\infty\).

Compute \(\displaystyle \lim_{x\to3}\dfrac{x-3}{2-\sqrt{3x-5}}\).

If we plug \(x=3\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). The denominator is a difference where one term is a square root. So we multiply the numerator and denominator by the conjugate, and only multiply out the denominator: \[\begin{aligned} \lim_{x\to3}&\dfrac{x-3}{2-\sqrt{3x-5}} =\lim_{x\to3}\left(\dfrac{x-3}{2-\sqrt{3x-5}}\right) \left(\dfrac{2+\sqrt{3x-5}}{2+\sqrt{3x-5}}\right) \\ &=\lim_{x\to3}\dfrac{(x-3)(2+\sqrt{3x-5})}{4-(3x-5)} \\ &=\lim_{x\to3}\dfrac{(x-3)(2+\sqrt{3x-5})}{9-3x} =\lim_{x\to3}\dfrac{2+\sqrt{3x-5}}{-3}=-\,\dfrac{4}{3} \end{aligned}\]

The same kind of trick can be extended to cube roots (and higher roots) as long as we remember an identity such as \[ (a-b)(a^2+ab+b^2)=a^3-b^3 \]

Compute \(\displaystyle \lim_{x\to2}\dfrac{2-\sqrt[\scriptstyle 3]{4x}}{x-2}\).

If we plug \(x=2\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). The numerator is a difference where one term is a cube root. So we multiply the numerator and denominator by \(4+2(4x)^{1/3}+(4x)^{2/3}\): \[\begin{aligned} \lim_{x\to2}&\dfrac{2-\sqrt[\scriptstyle 3]{4x}}{x-2} \\ &=\lim_{x\to2}\left(\dfrac{2-\sqrt[\scriptstyle 3]{4x}}{x-2}\right) \left(\dfrac{4+2(4x)^{1/3}+(4x)^{2/3}}{4+2(4x)^{1/3}+(4x)^{2/3}}\right) \\ &=\lim_{x\to2}\dfrac{8-4x}{(x-2)(4+2(4x)^{1/3}+(4x)^{2/3})} \\ &=\lim_{x\to2}\dfrac{-4}{4+2(4x)^{1/3}+(4x)^{2/3}} \\ &=\dfrac{-4}{4+2(8)^{1/3}+(8)^{2/3}} =\dfrac{-4}{4+2\cdot2+4}=-\,\dfrac{1}{3} \end{aligned}\]

Compute \(\displaystyle \lim_{u\to0}\dfrac{\sqrt{4+u}-\sqrt{4-u}}{u}\).

\(\displaystyle \lim_{u\to0}\dfrac{\sqrt{4+u}-\sqrt{4-u}}{u}=\dfrac{1}{2}\)

If we plug \(u=0\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). The numerator is a difference of square roots. So we multiply the numerator and denominator by the conjugate: \[\begin{aligned} \lim_{u\to0}&\dfrac{\sqrt{4+u}-\sqrt{4-u}}{u} \\ &=\lim_{u\to0}\left(\dfrac{\sqrt{4+u}-\sqrt{4-u}}{u}\right) \left(\dfrac{\sqrt{4+u}+\sqrt{4-u}}{\sqrt{4+u}+\sqrt{4-u}}\right) \\ &=\lim_{u\to0}\dfrac{(4+u)-(4-u)}{u(\sqrt{4+u}+\sqrt{4-u})} \\ &=\lim_{u\to0}\dfrac{2u}{u(\sqrt{4+u}+\sqrt{4-u})} \\ &=\lim_{u\to0}\dfrac{2}{\sqrt{4+u}+\sqrt{4-u}}=\dfrac{1}{2} \end{aligned}\]