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Wize University Calculus 1 Textbook > Limits

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}}

Extra Practice
Limits: Multiplying by the Conjugate
lim⁡x→∞x2+2x−x2+1\lim_{x\rightarrow\infty}\sqrt{x^2+2x}-\sqrt{x^2+1}
Limit Techniques: Multiply by the Conjugate
lim⁡x→π21−sin⁡xcos⁡x\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}\frac{1-\sin{x}}{\cos{x}}
∞ - ∞ Limits with Square Root
lim⁡x→∞ x2+2x−x2−3x=\displaystyle\lim_{x\to\infty}\ \sqrt{x^2+2x}-\sqrt{x^2-3x}=
∞ - ∞ Limits with Square Root
lim⁡x→∞ x2+2x−x2−3x=\displaystyle\lim_{x\to\infty}\ \sqrt{x^2+2x}-\sqrt{x^2-3x}=
Evaluating limits: Multiplying by the Conjugate
Practice: Evaluating limits
Determine the value of lim⁡x →0 x+1−1−xx \lim_{x\ \rightarrow0}\ \frac{\sqrt{x+1}-\sqrt{1-x}}{x}\
Find the following limit: lim⁡x→3x2−9x−3\displaystyle\lim_{x\rightarrow3}\frac{x^2-9}{\sqrt{x}-\sqrt{3}}
Practice: Limit at Infinity with Roots
Q.\textbf{Q.} Evaluate the limit: lim⁡x→∞x2−2−x2+x\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2-2}-\sqrt{x^2+x}
Limit at Infinity with Roots
Evaluate lim⁡x→∞9x2+2x−3x\displaystyle \lim_{x\rightarrow\infty}\sqrt{9x^2+2x}-3x.
Limit with Unknown Coefficients
Find constants aa and bb such that lim⁡x→0ax+b−2x=1\displaystyle\lim_{x\rightarrow 0}\frac{\sqrt{ax+b}-2}{x}=1.
Square Roots: Limits by Multiplying by the Conjugate
Q.\textbf{Q.} Evaluate lim⁡x→2x+7−3x−2\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{x+7}-3}{x-2}
Limits by multiplying by the conjugate
lim⁡t→13+t−2t3−1\lim_{t\rightarrow1}\dfrac{\sqrt{3+t}-2}{t^3-1}
Evaluate lim⁡x→9x+7−42x−18\displaystyle\lim_{x\to9} \frac{\sqrt{x+7}-4}{2x-18}
l'Hopital's Rule
Evaluate
lim⁡x→∞x2+1−x\lim\limits_{x\to\infty} \sqrt{x^2+1}-x
Practice: Rationalizing Indeterminate Forms
Rationalizing: Evaluate lim⁡x→3x2−9x−3\lim_{x\to3}\limits\dfrac{x^2 - 9}{\sqrt x - \sqrt 3}
Limits: Multiplying Conjugates
Find constants a and b such that
lim⁡x→0ax+b−2x=1\lim\limits_{x\to 0} \frac{\sqrt{ax+b}-2}{x}=1
Limits: Multiplying Conjugates
Evaluate the limit:
lim⁡x→2x+7−3x−2\lim\limits_{x \to 2} \frac{\sqrt{x+7} - 3}{x-2}
The value of the limit lim⁡x→1x−1x2−3x+2\lim_{x\to1}\frac{\sqrt{x}-1}{x^2-3x+2} is
Limits by Multiplying by the Conjugate
The value of the limits lim⁡x→1x−1x2−4x+3\lim\limits_{x\rightarrow1}\frac{\sqrt{x}-1}{x^2-4x+3}
Determine the value of lim⁡x→24x−1−7x−2 \displaystyle \lim_{x \to 2} \frac{\sqrt{4x-1}-\sqrt{7}}{x-2}.
The value of the limit lim⁡x→1x−1x2−3x+2\lim_{x\to1}\frac{\sqrt{x}-1}{x^2-3x+2} is
Limit with Unknown Coefficients
Find constants aa and bb such that lim⁡x→0ax+b−2x=1\displaystyle\lim_{x\rightarrow 0}\frac{\sqrt{ax+b}-2}{x}=1.
Square Roots: Limits by Multiplying by the Conjugate
Q.\textbf{Q.} Evaluate lim⁡x→2x+7−3x−2\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{x+7}-3}{x-2}
Practice: Limit at Infinity with Roots
Q.\textbf{Q.} Evaluate the limit: lim⁡x→∞x2−2−x2+x\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2-2}-\sqrt{x^2+x}
Find the following limit: lim⁡x→3x2−9x−3\displaystyle\lim_{x\rightarrow3}\frac{x^2-9}{\sqrt{x}-\sqrt{3}}
The value of the limit lim⁡x→1x−1x2−3x+2\lim_{x\to1}\frac{\sqrt{x}-1}{x^2-3x+2} is
Limit at Infinity with Roots
Evaluate lim⁡x→∞9x2+2x−3x\displaystyle \lim_{x\rightarrow\infty}\sqrt{9x^2+2x}-3x.
Practice: Multiply by conjugate
Evaluatelim⁡x→−∞(x2+3x+x)\lim\limits_{x\rightarrow-\infin}(\sqrt{x^2+3x}+x)
Evaluate the limit
lim⁡x→23x−1−5x−2\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{3x-1}-\sqrt{5}}{x-2}
Evaluate lim⁡t→13+t−2t3−1\displaystyle \lim_{t\rightarrow 1}\frac{\sqrt{3+t}-2}{t^3-1}.
Determine the value of lim⁡x→∞4x4+4x2+1−2x2\displaystyle \lim_{x\rightarrow \infty} \sqrt{4x^4+4x^2+1}-2x^2.
🌶️ TOUGH! Evaluate the limit, lim⁡x→−∞x2−6x+x\displaystyle\lim_{x\rightarrow-\infty}\sqrt{x^2-6x}+x.
Find the following limit: lim⁡x→+∞x2−8x−x\displaystyle \lim_{x\rightarrow+\infty}\sqrt{x^2-8x}-x.
Evaluate lim⁡x→5x−5x2−4x−5\displaystyle \lim_{x\rightarrow 5}\frac{\sqrt{x}-\sqrt{5}}{x^2-4x-5}.
Limits: Combining techniques
Solve the limit lim⁡x→39x−6x−1x2−5−2\displaystyle \lim\limits_{x\to 3} \frac{\frac{9}{x}-\frac{6}{x-1}}{\sqrt{x^2-5}-2}.
lim⁡x→5x−5x2−4x−5\displaystyle\lim_{x\rightarrow5}\dfrac{\sqrt{x}-\sqrt{5}}{x^2-4x-5}
lim⁡x→∞x2+2x−x2+1\displaystyle\lim_{x\rightarrow\infty}\sqrt{x^2+2x}-\sqrt{x^2+1}
Evaluating limits: Multiplying by the Conjugate
Practice: Evaluating limits
Determine the value of lim⁡x →0 x+1−1−xx \lim_{x\ \rightarrow0}\ \frac{\sqrt{x+1}-\sqrt{1-x}}{x}\
∞ - ∞ Limits with Square Root
Practice: ∞ - ∞ Limits with Square Root
lim⁡x→∞ x2+2x−x2−3x=\displaystyle\lim_{x\to\infty}\ \sqrt{x^2+2x}-\sqrt{x^2-3x}=
Practice: $\frac{0}{0}$ Limit with Square Root
Practice: 0/0 Limit with Square Root
lim⁡x→0 xx+4−4−x=\displaystyle\lim_{x\to0}\ \frac{x}{\sqrt{x+4}-\sqrt{4-x}}=
Limits and Continuity
Evaluate the following limits:
a) lim⁡x→4x−4x−4\displaystyle \lim_{x \rightarrow 4} \frac{\sqrt{x} - \sqrt{4}}{x - 4}
b) lim⁡x→∞x2−5x+134x4+3x−21\displaystyle \lim_{x \rightarrow \infty} \frac{x^2 - 5x + 13}{\sqrt{4x^4 +3x - 21}}
Limit Techniques: Multiply by the Conjugate
lim⁡x→π21−sin⁡xcos⁡x\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}\frac{1-\sin{x}}{\cos{x}}
Limits by multiplying by the conjugate
lim⁡x→−∞(9x2−15x+3+3x)\displaystyle\lim_{x\rightarrow -\infty}\big(\sqrt{9x^2-15x+3}+3x\big)
Limits by multiplying by the conjugate
lim⁡x→5x−5x2−4x−5\displaystyle\lim_{x\rightarrow5}\dfrac{\sqrt{x}-\sqrt{5}}{x^2-4x-5}
Limits: Multiplying by the Conjugate
lim⁡x→∞x2+2x−x2+1\lim_{x\rightarrow\infty}\sqrt{x^2+2x}-\sqrt{x^2+1}
Limits by Multiplying by the Conjugate
lim⁡x→0x2+x−2−x\displaystyle\lim_{x\rightarrow0}\dfrac{x}{\sqrt{2+x}-\sqrt{2-x}}
Practice: L'hopital's rule
Q.\textbf{Q.} Evaluate the following limits.
1. lim⁡x→0+cos⁡xcot⁡x\displaystyle \lim_{x\rightarrow 0^+} \cos{x}^{\cot{x}}
2. lim⁡x→∞[x2+1−x]\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt{x^2+1}-x\right]