Question

Use Maclaurin series to evaluate the limits. \(\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{x^{2}}\)

Short answer

The limit is 4.

Step-by-step solution

Recall the Maclaurin Series for Sine

The Maclaurin series for \( \sin(x) \) is given by \( \sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \).

Use the Series to Find \( \sin(2x) \)

Substitute \( 2x \) for \( x \) in the series: \( \sin(2x) = 2x - \frac{(2x)^3}{6} + \frac{(2x)^5}{120} - \cdots \), which simplifies to \( \sin(2x) = 2x - \frac{8x^3}{6} + \frac{32x^5}{120} - \cdots \).

Square the Series Expression

Square the leading terms of the series expression: \( ( \sin(2x) )^2 = (2x - \frac{8x^3}{6} + \cdots )^2 \). To simplify, focus on the first term: \( ( \sin(2x) )^2 ≈ (2x)^2 = 4x^2 \) since higher-order terms will tend to zero faster.

Simplify the Limit Expression

Substitute the squared term back into the limit: \( \lim_{x \to 0} \frac{\sin^2(2x)}{x^2} = \lim_{x \to 0} \frac{4x^2}{x^2} = 4 \).

Key concepts

Limit Evaluation

Assessing limits using series expansions can simplify seemingly complicated expressions. In this case, the limit \(\lim \_{x \rightarrow 0} \frac{\sin \^{2} 2x}{x\^{2}}\) is easier to handle when using the Maclaurin series. The Maclaurin series for \[\sin(x)\] enables us to break down the function into an infinite sum of terms. We then substitute these terms into the limit function. By focusing on the dominant terms (those with the lowest powers), we can simplify our calculations. This approach leverages the behavior of polynomial terms, where higher power terms tend to zero faster as \[x\] approaches zero. Thus, concentrating on the first few terms can provide an accurate limit result.

Series Expansion

A series expansion is a way to express functions as infinite sums of simpler terms. The Maclaurin series is a type of Taylor series expansion at \[x = 0\]. For \[\sin(x)\], the series is \[\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\]. When we substitute \[2x\] for \[x\], we get: \[\sin(2x) = 2x - \frac{(2x)^3}{6} + \frac{(2x)^5}{120} - \cdots\]. This simplifies to \[\sin(2x) = 2x - \frac{8x^3}{6} + \frac{32x^5}{120} - \cdots\]. Series expansion is instrumental in calculus as it provides insights into the function’s behavior near a specific point. It also makes complex trigonometric or exponential functions more approachable by converting them into polynomial forms.

Trigonometric Functions

Trigonometric functions such as \[\sin(x)\] and \[\cos(x)\] have specific series representations that are very useful in calculus. The Maclaurin series for \[\sin(x)\] and \[\cos(x)\] are:
- \[\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\]
- \[\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\]
These series help analyze the behaviors of these functions near zero. Using series expansions, we can derive limits, solve differential equations, and approximate function values. For instance, squaring \[\sin(2x)\] in our problem, primarily concerns terms up to \[x^2\] in order to simplify the limit evaluation accurately. Understanding these expansions can help students tackle many problems in calculus more effectively.