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Wize AP Calculus (AB) Textbook > Limits & Continuity

Computing Limits by Multiplying by the Conjugate

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Computing Limits by Multiplying by the Conjugate

When dealing with limits involving square roots or certain trig identities, sometimes the correct idea is to multiply by a "special 1". This special one is the conjugate over itself.

Conjugates

The conjugate of a+ba+b is a−ba-b.

The conjugate of a−ba-b is a+ba+b.

Notice that multiplying by a conjugate gives us the difference of squares formula:

(a+b)(a−b)=a2−b2\boxed{(a+b)(a-b)=a^2-b^2}

Quick Examples

  1. x−y=(x−y)x+yx+y=x−yx+y\displaystyle\sqrt{x}-\sqrt{y}=(\sqrt{x}-\sqrt{y})\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\frac{x-y}{\sqrt{x}+\sqrt{y}}

  1. 11−cos⁡x=1(1−cos⁡x)1+cos⁡x(1+cos⁡x)=1+cos⁡x1−cos⁡2x=1+cos⁡xsin⁡2x\displaystyle\frac{1}{1-\cos{x}}=\frac{1}{(1-\cos{x})}\frac{1+\cos{x}}{(1+\cos{x})}=\frac{1+\cos{x}}{1-\cos^2{x}}=\frac{1+\cos{x}}{\sin^2{x}}






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Example: Multiply by the Conjugate

Find the following limit

lim⁡x→3x2−9x−3\displaystyle\lim_{x\rightarrow3}\frac{x^2-9}{\sqrt{x}-\sqrt{3}}



lim⁡x→3x2−9x−3=lim⁡x→3x2−9x−3x+3x+3=lim⁡x→3(x−3)(x+3)(x+3)x−3\lim_{x \to 3}\dfrac{x^2-9}{\sqrt{x}-\sqrt{3}}=\lim_{x \to 3}\dfrac{x^2-9}{\sqrt{x}-\sqrt{3}}\dfrac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}\\ \text{} \\=\lim_{x \to 3}\dfrac{(\cancel{x-3})(x+3)(\sqrt{x}+\sqrt{3})}{\cancel{x-3}}

=lim⁡x→3(x+3)(x+3)=6(23)=123=\lim_{x \to 3}(x+3)(\sqrt{x}+\sqrt{3})\\ \text{} \\=6(2\sqrt{3})\\ \text{} \\=12\sqrt{3}


Find the following limit

lim⁡x→π21−sin⁡xcos⁡x\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}\frac{1-\sin{x}}{\cos{x}}