Evaluating Limits Using the Rationalizing Technique
Core Concepts
Learn to evaluate limits involving radicals that result in the \(\frac{0}{0}\) indeterminate form. This stack focuses on using the conjugate to manipulate the expression and reveal a solution.
Evaluating Limits Using the Rationalizing Technique
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When is the rationalizing technique the most appropriate method for evaluating a limit?
This technique is ideal when direct substitution leads to \(\frac{0}{0}\) and the expression involves a square root, typically in the form of a sum or difference like \(\sqrt{a} - b\).
What is the conjugate of the expression \(\sqrt{a} + b\)?
The conjugate is \(\sqrt{a} - b\). To find the conjugate, you simply change the sign between the two terms.
What is the algebraic result of multiplying an expression like \((\sqrt{a} - \sqrt{b})\) by its conjugate, \((\sqrt{a} + \sqrt{b})\)?
The result is \(a - b\). This is based on the difference of squares formula, \((x-y)(x+y) = x^2 - y^2\), which eliminates the square roots.
$$\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$$
Multiply by the conjugate \(\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\). This gives $$\lim_{x \to 0} \frac{(x+1)-1^2}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{2}$$
When rationalizing the numerator of a fraction, should you multiply out the terms in the denominator?
Generally, no. Leave the denominator in factored form. The goal is to create a term in the simplified numerator that will cancel with a factor in the original denominator.
$$\lim_{x \to 4} \frac{x-4}{\sqrt{x}-2}$$
Here, the radical is in the denominator. Multiply by its conjugate, \(\frac{\sqrt{x}+2}{\sqrt{x}+2}\). This yields $$\lim_{x \to 4} \frac{(x-4)(\sqrt{x}+2)}{x-4} = \lim_{x \to 4} (\sqrt{x}+2) = \sqrt{4}+2 = 4$$
Can the rationalizing technique be applied if the radical expression is not a square root (e.g., a cube root)?
Yes, but it's more complex. For cube roots, you would use the sum or difference of cubes formulas (e.g., \((a-b)(a^2+ab+b^2) = a^3-b^3\)) to find the appropriate 'conjugate' that eliminates the cube root.
Evaluate the limit which represents the derivative of \(f(x)=\sqrt{x}\) at \(x=9\): $$\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}$$
Multiply numerator and denominator by the conjugate \(\sqrt{9+h}+3\). This gives $$\lim_{h \to 0} \frac{(9+h)-9}{h(\sqrt{9+h}+3)} = \lim_{h \to 0} \frac{h}{h(\sqrt{9+h}+3)} = \lim_{h \to 0} \frac{1}{\sqrt{9+h}+3} = \frac{1}{6}$$
