Evaluate The Following Limit Using Taylor Series

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

When faced with a challenging limit that results in an indeterminate form like 0/0 or ∞/∞, most students immediately reach for L’Hôpital’s Rule. This method leverages the fundamental idea that any smooth function can be approximated by a polynomial near a point of interest. In practice, by replacing functions with their Taylor polynomials, we transform a difficult limit into a straightforward algebraic simplification of polynomial ratios. While effective, this method can become cumbersome with complex functions, often requiring multiple differentiations that lead to more complicated expressions. Worth adding: a more elegant and often simpler approach exists: evaluating limits using Taylor series expansions. This technique not only provides the answer but also offers deep insight into the local behavior of functions.

Why Use Taylor Series for Limits?

The primary strength of the Taylor series approach lies in its ability to systematically approximate a function to a desired degree of accuracy. The Taylor series of a function f(x) around a is: f(x) = f(a) + f'(a)(x-a) + (f''(a)/2!Now, for limit evaluation, we are concerned with the behavior as the variable approaches a specific point, say x → a`. )(x-a)² + (f'''(a)/3!)(x-a)³ + ...

For the purpose of calculating a limit, we only need to expand the numerator and denominator to a sufficient order—just enough terms to resolve the indeterminate form. Lower-order terms will cancel out or become negligible. Think about it: this contrasts with L’Hôpital’s Rule, which can sometimes require many iterations and produce unwieldy derivatives. To build on this, the Taylor method works without friction for limits at infinity by using expansions like sin(x) ~ x for small x, or by substituting t = 1/x and expanding around t=0 Less friction, more output..

Step-by-Step Methodology

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

1. Identify the Point of Expansion: The limit point a (which could be 0, a finite number, or ∞) dictates where you center the Taylor series. For x→0, use the Maclaurin series (a special case of Taylor series centered at 0).
2. Expand the Numerator and Denominator: Write the Taylor series for both f(x) and g(x) around x = a. Include terms up to the smallest power that will not cancel out completely. A good rule of thumb: expand until the first non-zero term in both numerator and denominator after substitution is of the same order.
3. Substitute and Simplify: Replace f(x) and g(x) in the limit expression with their polynomial approximations. Cancel out any common factors of (x-a) from the numerator and denominator.
4. Evaluate the Simplified Limit: The resulting expression will be a simple ratio of constants and powers of (x-a). As x→a, any term with a positive power of (x-a) will vanish, leaving a finite, easily computable result.
5. Verify Sufficiency: If the result is still indeterminate (e.g., 0/0 after initial cancellation), return to step 2 and include the next higher-order term in your expansions.

This method is not merely a trick; it is grounded in the rigorous definition of a limit and the concept of asymptotic equivalence. If f(x) ~ c*(x-a)^k and g(x) ~ d*(x-a)^m as x→a, then f(x)/g(x) ~ (c/d)*(x-a)^{k-m}. The limit depends on the exponents k and m and the leading coefficients c and d And it works..

Worked Examples: From Simple to Complex

Example 1: The Classic 0/0 Form Evaluate lim_{x→0} (sin(x) - x) / x³.

• Step 1 & 2: Expand sin(x) around 0: sin(x) = x - x³/3! + x⁵/5! - .... That's why, sin(x) - x = -x³/6 + x⁵/120 - ....
• Step 3: Substitute: ( -x³/6 + higher order terms ) / x³ = -1/6 + (x²/120) - ....
• Step 4: As x→0, all higher-order terms vanish. The limit is -1/6.
• Why it’s better: Applying L’Hôpital’s Rule here would require three differentiations, each producing more complex trigonometric derivatives. The Taylor method is a one-step conceptual process.

Example 2: A More Involved Trigonometric Limit Evaluate lim_{x→0} (1 - cos(x)) / (x sin(x)).

• Step 1 & 2: Use known expansions: cos(x) = 1 - x²/2! + x⁴/4! - ..., so 1 - cos(x) = x²/2 - x⁴/24 + .... Also, sin(x) = x - x³/6 + ..., so x sin(x) = x² - x⁴/6 + ....
• Step 3: The ratio becomes (x²/2 - x⁴/24 + ...) / (x² - x⁴/6 + ...). Factor x²: [1/2 - x²/24 + ...] / [1 - x²/6 + ...].
• Step 4: As x→0, this simplifies to (1/2) / 1 = 1/2. The limit is 1/2.

Example 3: Limit at Infinity (Using Substitution) Evaluate lim_{x→∞} x * sin(1/x).

• Step 1: Let t = 1/x. As x→∞, t→0⁺. The limit becomes `lim_{t→0⁺} (1/t) * sin(t) = lim_{t→0⁺}