Evaluate The Following Limit Using Taylor Series

Evaluating Limits Using Taylor Series: A Powerful Alternative to L’Hôpital’s Rule

When faced with a stubborn limit that results in an indeterminate form like 0/0 or ∞/∞, most students immediately reach for L’Hôpital’s Rule. While effective, this method can become cumbersome, requiring multiple differentiations and often leading to algebraic complexity. A more elegant, insightful, and often simpler approach exists: evaluating limits using Taylor series. This technique leverages the power of polynomial approximations to transform seemingly intractable limit problems into straightforward arithmetic. By replacing transcendental functions with their polynomial expansions around the point of interest, we can dissect the behavior of a function with surgical precision, revealing its limit directly through dominant terms.

Why Taylor Series Excel at Limit Evaluation

Traditional algebraic manipulation and L’Hôpital’s Rule are essentially brute-force methods. They work, but they don’t always provide deep insight into why a limit takes a particular value. Taylor series, however, provide a local approximation of a function as an infinite sum of terms calculated from its derivatives at a single point. For limit evaluation, we don’t need the entire infinite series; we only need enough terms to capture the behavior of the function near the point of interest.

The core idea is this: if you have a limit lim_{x→a} f(x)/g(x) that is indeterminate, you can replace f(x) and g(x) with their Taylor polynomials (the finite truncations of their Taylor series) centered at x = a. You then perform the division and simplify. The limit as x→a is simply the ratio of the first non-zero terms in these polynomials, provided the orders of the zeroes (or poles) are correctly matched. This method systematically avoids the repetitive differentiation of L’Hôpital’s and makes the dominant balance between numerator and denominator explicitly clear.

The Step-by-Step Methodology

To evaluate lim_{x→a} f(x)/g(x) using Taylor series, follow this structured protocol:

1. Identify the Indeterminate Form and Center: Confirm the limit is indeterminate (0/0, ∞/∞, etc.). The natural center for your Taylor expansions is the point x = a to which the limit is approaching. If the limit is as x→0, you will use Maclaurin series (Taylor series centered at 0).

2. Expand the Functions: Write the Taylor series for f(x) and g(x) around x = a. You must expand both the numerator and the denominator. The critical rule is to expand each function to a sufficiently high order (degree) so that you capture the first non-zero term in their difference from their value at a. A practical guideline: expand each until you have at least two non-zero terms beyond the constant term, or until the orders of the leading non-zero terms in the numerator and denominator are clear.

3. Substitute and Simplify: Replace f(x) and g(x) in your limit expression with their truncated polynomial expansions. Perform the algebraic division. You will often have a polynomial in the numerator divided by a polynomial in the denominator.

4. Evaluate the Limit: As x→a, all terms containing factors of (x-a) will vanish. The limit is therefore determined by the ratio of the constant terms (if the polynomials are of degree 0) or, more commonly, by the ratio of the coefficients of the lowest power of (x-a) present in both the numerator and denominator after simplification.

Crucial Insight: If after substitution the numerator has a leading term of order (x-a)^m and the denominator has a leading term of order (x-a)^n, then:

• If m > n, the limit is 0.
• If m = n, the limit is (coefficient of (x-a)^m in num) / (coefficient of (x-a)^n in den).
• If m < n, the limit is ±∞ (depending on signs).

Worked Examples: From Simple to Sophisticated

Example 1: The Classic (sin x)/x as x→0 This is the foundational limit. Using the Maclaurin series for sin x: sin x = x - x³/3! + x⁵/5! - ... Our limit becomes: lim_{x→0} (x - x³/6 + ...) / x = lim_{x→0} (1 - x²/6 + ...) All higher-order terms vanish as x→0. The limit is 1, from the constant term 1.

Example 2: A More Complex Case (e^x - 1 - x) / (x sin x) as x→0 Here, both numerator and denominator approach 0. We need expansions to at least the x² term.

• e^x = 1 + x + x²/2! + x³/3! + ... So, e^x - 1 - x = x²/2 + x³/6 + ...
• sin x = x - x³/3! + ... So, x sin x = x*(x - x³/6 + ...) = x² - x⁴/6 + ... The limit becomes: `lim_{x→0} (x²/2 + x³/6 + ...) / (x² - x⁴/6 + ...