Evaluating Limits by Simplifying Complex Fractions
Core Concepts
Focus on the technique for solving limits that involve complex fractions (fractions within a fraction). The primary method is to find a common denominator to simplify the expression before evaluating the limit.
Evaluating Limits by Simplifying Complex Fractions
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What is a complex fraction, and when does it pose a problem for evaluating limits?
A complex fraction is a fraction that has one or more fractions in its numerator, denominator, or both. It becomes a problem when direct substitution results in the indeterminate form \(\frac{0}{0}\).
What is the primary strategy for simplifying a complex fraction when evaluating a limit?
Combine the terms in the numerator (or denominator) into a single fraction by finding a common denominator. Then, simplify the resulting larger fraction, often by multiplying by the reciprocal of the denominator.
Simplify the numerator by finding a common denominator of \(2(x+2)\): $$\lim_{x \to 0} \frac{\frac{2 - (x+2)}{2(x+2)}}{x} = \lim_{x \to 0} \frac{-x}{2x(x+2)}$$ Cancel \(x\) to get $$\lim_{x \to 0} \frac{-1}{2(x+2)} = -\frac{1}{4}$$
What is an alternative method to simplifying a complex fraction, besides finding a common denominator for the inner fractions?
Multiply the entire numerator and the entire denominator of the main fraction by the least common multiple (LCM) of the denominators of the 'mini' fractions. This often clears the inner fractions in one step.
Using the LCM method, solve $$\lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x-3}$$
The LCM of the inner denominators (\(x\) and \(3\)) is \(3x\). Multiply the top and bottom by \(3x\): $$\lim_{x \to 3} \frac{3x(\frac{1}{x} - \frac{1}{3})}{3x(x-3)} = \lim_{x \to 3} \frac{3 - x}{3x(x-3)} = \lim_{x \to 3} \frac{-(x-3)}{3x(x-3)} = \lim_{x \to 3} \frac{-1}{3x} = -\frac{1}{9}$$
After simplifying a complex fraction, what analytical technique is almost always required next?
The factoring and canceling technique. Simplifying the complex fraction is designed to reveal a common factor between the numerator and denominator that can be canceled.
Evaluate the limit representing the derivative of \(f(x) = \frac{1}{x}\): $$\lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$$
The common denominator in the numerator is \(x(x+h)\). This simplifies to $$\lim_{h \to 0} \frac{x - (x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-h}{h x(x+h)}$$ Cancel \(h\) to get $$\lim_{h \to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^2}$$
