Problem 32 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow+\infty}[\cos (2 / x)]^{x^{2}}$$

Step-by-Step Solution

Answer

The limit is \( e^{-2} \).

Step 1: Understand the Limit Expression

We want to find \( \lim_{x \rightarrow +\infty} [\cos (2 / x)]^{x^2} \). This expression involves a cosine function where the angle approaches zero as \( x \) becomes very large, and it is raised to the power of \( x^2 \).

Step 2: Simplify the Inner Cosine Term

As \( x \rightarrow +\infty \), the term \( \frac{2}{x} \rightarrow 0 \). So, we need to evaluate \( \cos(\frac{2}{x}) \approx \cos(0) = 1 \). Thus, \( \cos(\frac{2}{x}) \approx 1 \) when \( x \) is very large.

Step 3: Use Exponential Logarithm Form

Express \( [\cos(2/x)]^{x^2} \) in terms of the exponential function: \[ y = [\cos(2/x)]^{x^2} \] Take the logarithm: \[ \ln(y) = x^2 \ln(\cos(2/x)) \]

Step 4: Expand Logarithm using Taylor Series

Using the Taylor series expansion for \( \ln(\cos(\theta)) \) when \( \theta \approx 0 \) gives: \[ \ln(\cos(2/x)) \approx -\frac{1}{2} \left(\frac{2}{x}\right)^2 = -\frac{2}{x^2} \] Substitute this into the equation from Step 3: \[ \ln(y) = x^2 \cdot\left( -\frac{2}{x^2} \right) = -2 \]

Step 5: Evaluate the Limit of the Exponential Function

With \( \ln(y) = -2 \), we apply the exponential function to find \( y \): \[ y = e^{-2} \] Therefore, the limit is \( e^{-2} \).

Step 6: Conclusion

Combining all the computations, we find \[ \lim_{x \rightarrow +\infty} [\cos(2/x)]^{x^2} = e^{-2} \]

Key Concepts

Taylor Series

The Taylor series is a powerful way to represent functions as an infinite sum of terms. Each term is calculated from the derivative of the function at a single point. It can approximate complex functions using polynomials, which makes calculations easier.
For example, if you have a function like \( f(x) \), the Taylor series expansion around a point \( a \) is:

• \( f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \)

In the given problem, we used the Taylor series to approximate \( \ln(\cos(\theta)) \) for small values of \( \theta \). Since \( \theta = \frac{2}{x} \) becomes small as \( x \to +\infty \), we applied:

• \( \ln(\cos(\theta)) \approx -\frac{1}{2}\theta^2 \)

This approximation allowed us to simplify the logarithmic expression, helping in solving the limit problem more efficiently.

Exponential Functions

Exponential functions are mathematically expressed as \( e^x \), where \( e \) is the mathematical constant approximately equal to 2.71828. This kind of function grows rapidly and has unique properties in calculus.
They are crucial in limit problems where expressions involve powers or logarithms. In step 5 of the solution, after simplifying using the Taylor series and the properties of logarithms, we reached a situation where:\[ \ln(y) = -2 \]
By applying the exponential function to both sides, we get:

• \( y = e^{-2} \)

This involved transitioning from the logarithmic form to the exponential form, which is common in solving limits. The exponential function really helps convert complex log expressions into more manageable forms.

Trigonometric Limits

Trigonometric functions like sine, cosine, and tangent often appear in calculus problems, especially limits. Understanding how these functions behave as they approach certain values is key to solving complex limit problems.
For instance, the cosine function \( \cos(\theta) \) becomes very helpful when the angle \( \theta \) is small. In the original exercise, as \( x \to +\infty \), the angle approximates zero due to \( \cos(2/x) \to \cos(0) = 1 \).
This allows us to approximate \( \cos(2/x) \approx 1 - \frac{1}{2}(2/x)^2 \) when using Taylor series. Such approximations simplify calculating trigonometric limits involving powers, making the problem easier to solve. Recognizing these patterns and small angle approximations is often essential in calculus.