Problem 64 - Single Variable Calculus

Question

Evaluate \(\lim _{x \rightarrow 0} \frac{\cos x^{2}-1+\frac{1}{2} x^{4}}{x^{8}}\). Hint: Use the Maclaurin series representation of \(\cos x^{2}\).

Step-by-Step Solution

Answer

The Maclaurin series representation of \(\cos x^2\) is \(1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \cdots\). So the given expression simplifies to \(\frac{x^4(- \frac{1}{2!} + \frac{1}{2} + \frac{x^4}{4!} - \cdots)}{x^8}\). After canceling common factors and evaluating the limit as x approaches 0, we find that \(\lim _{x \rightarrow 0} \frac{\cos x^{2}-1+\frac{1}{2} x^{4}}{x^{8}} = 0\).

Step 1: Recall the Maclaurin series for cosine function

To find the Maclaurin series representation of \(\cos x^2\), we recall the standard Maclaurin series for the cosine function: \(\cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\) Now, we replace \(x\) with \(x^2\):

Step 2: Maclaurin series for \(\cos x^2\)

Replace \(x\) with \(x^2\) to get the Maclaurin series representation for \(\cos x^2\): \(\cos x^2 = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{4n} = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \cdots\)

Step 3: Substitute the Maclaurin series into the expression

Now, we substitute the Maclaurin series representation for \(\cos x^2\) into the expression in the problem: \(\frac{\cos x^2 - 1 + \frac{1}{2}x^4}{x^8} = \frac{(1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots) - 1 + \frac{1}{2}x^4}{x^8}\) And then simplify the expression:

Step 4: Simplify the expression

\(\frac{- \frac{x^4}{2!} + \frac{1}{2}x^4 + \frac{x^8}{4!} - \cdots}{x^8}\) Now, we factor out the \(x^4\) from the numerator: \(\frac{x^4\left(- \frac{1}{2!} + \frac{1}{2} + \frac{x^4}{4!} - \cdots\right)}{x^8}\)

Step 5: Cancel the common factors and evaluate the limit

Notice that we can cancel out the \(x^4\) in both the numerator and the denominator: \(\lim_{x \rightarrow 0} \frac{- \frac{1}{2!} + \frac{1}{2} + \frac{x^4}{4!} - \cdots}{x^4}\) Now, let's take the limit as x approaches 0: \(\lim_{x \rightarrow 0} \left(- \frac{1}{2!} + \frac{1}{2} + \frac{x^4}{4!} - \cdots\right) = - \frac{1}{2} + \frac{1}{2}\) The result is:

Step 6: The final answer

The limit of the given function as x approaches 0 is: \(\lim _{x \rightarrow 0} \frac{\cos x^{2}-1+\frac{1}{2} x^{4}}{x^{8}} = 0\)

Key Concepts

Maclaurin Series Representation

Understanding the Maclaurin series is crucial for simplifying complex functions into infinite sums of polynomial terms. It's a special case of the Taylor series centered at zero. The idea is to approximate more complex functions with a polynomial that has the same derivatives at a single point, in this case, the point is zero.

When we express the cosine function as a Maclaurin series, it takes the general form: \[\begin{equation} \cos(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}, \end{equation}\]
where the exclamation mark denotes a factorial, which is the product of all positive integers less than or equal to the number. For the cosine function, the series contains only even powers of \(x\), with alternating positive and negative signs, reflecting the periodicity and symmetry of the cosine wave.

Limits in Calculus

In calculus, a limit is the value that a function or sequence 'approaches' as the input or index approaches some value. Limits are essential for defining the fundamental concepts of calculus, such as derivatives and integrals. They also come in handy when evaluating the behavior of functions near a point of interest, something that cannot be directly observed.

The limit we are dealing with in the textbook exercise involves the function as \(x\) approaches zero, and using the established Maclaurin series for the cosine function helps us to evaluate this limit. By expanding the function into a form that cancels out higher-order terms, we're left with a simple expression whose limit can be more easily seen.

Cosine Function

The cosine function is one of the primary trigonometric functions and describes the relationship between the angles and lengths of a right triangle. Geometrically, in the unit circle, it represents the x-coordinate of a point determined by an angle \(x\). Its wave-like graph is essential when modeling periodic phenomena such as sound and light waves.

Mathematically, for real numbers \(x\), the cosine function also exhibits even symmetry and has a Maclaurin series that consists of even powers of \(x\) only. This is an outcome of the fact that cosine is an even function, meaning that \(\cos(-x) = \cos(x)\). These properties are key to simplifying expressions when cosine is involved and we’re taking limits around the point \(x=0\).

Simplifying Mathematical Expressions

Simplifying mathematical expressions is a practice to make them easier to understand or to allow for further analysis. In our context, simplification involves factorizing, combining like terms, and canceling out common factors. By doing so, we can reduce the complexity of the problem and often reveal the nature of limits.

For the given exercise, after substituting the Maclaurin series into the expression, we simplify the numerator by combining like terms and factoring out a common \(x^4\). We then cancel it against the \(x^8\) in the denominator, which conveniently reduces the expression to one where the limit can be directly applied. Hence, simplification is a powerful tool, not only for clearing the clutter but also for shedding light on the otherwise hidden limits of functions.