Limits Evaluate the following limits using Taylor series.
lim ₓ→₀ (sin 2x)/x

Verified step by step guidance

1. Recall that the Taylor series expansion of \( \sin x \) around \( x = 0 \) is given by \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \).
2. To find the Taylor series for \( \sin 2x \), substitute \( 2x \) into the series for \( \sin x \), giving \( \sin 2x = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots \).
3. Write the expression \( \frac{\sin 2x}{x} \) using the series expansion: \( \frac{2x - \frac{(2x)^3}{3!} + \cdots}{x} \).
4. Simplify the fraction by dividing each term in the numerator by \( x \), resulting in \( 2 - \frac{(2x)^3}{3! x} + \cdots \).
5. Evaluate the limit as \( x \to 0 \) by noting that all terms containing \( x \) vanish, leaving the constant term as the limit.

Key Concepts

Limits

A limit describes the value that a function approaches as the input approaches a certain point. Understanding limits is fundamental in calculus for analyzing behavior near specific points, especially when direct substitution leads to indeterminate forms like 0/0.

Taylor Series

A Taylor series represents a function as an infinite sum of terms calculated from its derivatives at a single point. It approximates functions near that point, allowing complex expressions to be simplified and limits to be evaluated more easily.

Evaluating Limits Using Taylor Series

By expanding functions into their Taylor series near the limit point, indeterminate forms can be resolved by canceling terms or simplifying expressions. This method transforms complicated limits into algebraic ones, making it easier to find the limit value.