1.6 Determining Limits Using Algebraic Manipulation
Keywords
Indeterminate Form 不定式 • Factoring 因式分解
Conjugates 共轭 • Trigonometric Identities 三角恒等式 • Rationalizing 有理化
The Indeterminate Form \(\frac{0}{0}\)
When evaluating \(\lim\limits_{x \to c} \frac{f(x)}{g(x)}\), if direct substitution results in \(\frac{0}{0}\), the limit is in an indeterminate form. This does not mean the limit does not exist; rather, it means the limit is "hidden" and we must use algebra to reveal it.
核心要点：处理不定式
If direct substitution yields \(\frac{0}{0}\), you must rearrange the expression into an equivalent form. The goal is usually to "cancel out" the term causing the zero in the denominator.
Strategy 1: Factoring and Dividing
This is the most common technique for rational functions. If \(f(c)=0\) and \(g(c)=0\), then \((x-c)\) is a factor of both the numerator and denominator.
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Factor the numerator and denominator completely.
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Cancel the common factor \((x-c)\).
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Substitute \(x=c\) into the remaining expression.
Example: A Hole in the Graph
Evaluate \(\lim\limits_{x \to 1} \frac{x^2 - 1}{x - 1}\). $$ \frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1} = x+1 $$ Now, substitute \(x=1\): \(1 + 1 = 2\).
Strategy 2: Rationalization (Conjugates)
If the expression contains a square root, multiplying by the conjugate can help clear the indeterminate form. Recall that the conjugate of \((a - \sqrt{b})\) is \((a + \sqrt{b})\).
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Identify the radical expression.
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Multiply both the numerator and denominator by the conjugate of that expression.
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Simplify using the difference of squares: \((a-b)(a+b) = a^2 - b^2\).
Example:
Evaluate \(\lim\limits_{x \to 0} \frac{\sqrt{x+4}-2}{x}\). Multiplying by \(\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\):
Substituting \(x=0\): \(\frac{1}{\sqrt{4}+2} = \frac{1}{4}\).
Strategy 3: Trigonometric Simplification
Sometimes we use trigonometric identities to rewrite the expression before substituting.
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Common Identities: \(\tan x = \frac{\sin x}{\cos x}\), \(\sin^2 x + \cos^2 x = 1\).
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Special Limits: (To be mastered in later sections, but helpful now) \(\lim\limits_{x \to 0} \frac{\sin x}{x} = 1\).
Extension: Removable Discontinuities
When we "cancel out" a factor like \((x-c)\) from the denominator, we are identifying a removable discontinuity (a hole). Algebraically, we are finding a new function that is identical to the original everywhere except at \(x=c\). Since the limit only cares about the approach and not the point itself, evaluating the new function gives us the correct limit for the original.