Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{x^{2}}$$

Understanding the Maclaurin series expansion for \(\text{sin}(x)\) is crucial for evaluating complex functions, as seen in our given exercise. The Maclaurin series is a special form of the Taylor series that expands a function around \(x = 0\). The series for \(\text{sin}(x)\) is:\[\text{sin}(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \text{higher-order terms}\]For small values of \(x\), higher-order terms become minimal. Hence, the approximation \( \text{sin}(x) \approx x \) works effectively. Applying this approximation simplifies calculations, making it especially useful for limit evaluation, as previously shown. It’s a powerful method allowing us to handle trigonometric functions with ease and accuracy in many calculus problems.If you think about \(\text{sin}(2x)\), applying the same principle where higher-order terms are neglected for small \(x\), you get:\[\text{sin}(2x) \approx 2x\]This simplification is essential for performing swift calculations in exercises requiring series expansion expertise.

When faced with evaluating limits, series approximations offer profound advantages. In our exercise, we evaluated \(\text{lim}_{x \to 0} \frac{\text{sin}^2(2x)}{x^2}\) using the Maclaurin series. Here's a breakdown of why this approach is effective:Firstly, recognizing that \( \text{sin}(2x) \approx 2x \) simplifies the initial trigonometric form into a basic polynomial. This leads to:\[ \frac{\text{sin}^2(2x)}{x^2} \approx \frac{(2x)^2}{x^2} = \frac{4x^2}{x^2} = 4\]Notice how initial complex expressions become manageable through such approximations. Clearly, applying these principles reveals the limit quickly.In the end, you found that \(\text{lim}_{x \to 0} \frac{\text{sin}^2(2x)}{x^2} \)= 4, showing the value stays constant as \(x \) approaches zero.Breaking down limits using series approximations not only simplifies the steps but also ensures accuracy, providing an excellent strategy for handling complex limits involving trigonometric functions.

Basic series approximations transform challenging mathematics into simple steps. By neglecting higher-order terms for small \(x\), the Maclaurin series gives a close approximation of the actual function value.Take \(\text{sin}(x)\) as an example, approximated by \(\text{sin}(x) \approx x \) for small \( x \). For \( \text{sin}(2x) \), it becomes \(\text{sin}(2x) \approx 2x \), significantly simplifying the calculation.These approximations make it easier to work through exercises that involve functions whose exact values are complicated to derive manually.Here's why such approximations are useful:

• They allow rapid mental calculations.
• They simplify limits involving trigonometric functions.
• They facilitate deeper understanding of function behavior near zero.

In essence, basic series approximations provide valuable tools in calculus, turning seemingly complex problems into straightforward solutions.