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Find values of \(a\) and \(b\) for which $$\lim _{x \rightarrow 0} \frac{\cos a x-b}{2 x^{2}}=-4$$

Short Answer

Expert verified
The values that solve the limit are \(a = 2i\) or \(a = -2i\) and \(b = 0\).

Step by step solution

01

Simplify the function inside the limit

Let's rewrite the function inside the limit by re-arranging terms using the identity that \(\cos 0 = 1\). This results in: \[\lim_{x \to 0} \frac{1 - \cos ax + b - 1}{2x^2}\] which further simplifies to \[\lim_{x \to 0} \frac{1 - \cos ax}{2x^2} - \frac{b}{2x^2}\].
02

Use limit formula of \(1-\cos ax\)

The limit \(\lim_{x \to 0} \frac{1 - \cos ax}{x^2}\) is a standard limit, and its value is \(\frac{a^2}{2}\). Thus, \[\lim_{x \to 0} \frac{1 - \cos ax}{2x^2} = \frac{a^2}{4}\]. Therefore, the given limit equation can be written as \[\frac{a^2}{4} - \lim_{x \to 0} \frac{b}{2x^2} = -4\].
03

Determine the limit of \(b/2x^2\)

\(\lim_{x \to 0} \frac{b}{2x^2}\) is undefined, and for the overall limit to equal -4, this term needs to be 0. Therefore, \(b\) must equals 0.
04

Solve for \(a\)

Let's solve \(\frac{a^2}{4} = -4\). To do this, multiply both sides by 4 and take the square root. Solving it gives \(a = 2i\) or \(a = -2i\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits and Continuity
Understanding the concept of limits is fundamental in calculus. The limit of a function describes the behavior of the function as the input approaches a certain value. In our exercise, we are asked to find the values of two constants, \(a\) and \(b\), such that the limit of a given function as \(x\) approaches 0 is \(-4\). To ensure that the limit exists and the function is continuous at that point, all terms within the function that could become undefined or infinitely large as \(x\) approaches 0 must be addressed.

Continuity of a function at a certain point means the function's limit at that point is equal to the function's value at that point. In cases where a function is not continuous, like a term involving \(\frac{1}{x^2}\) as \(x\) approaches 0, the respective term must simplify to a constant or equal zero for the limit to exist. In our exercise, we use this principle to conclude that \(b\) must be zero to prevent an undefined term in the limit.
Trigonometric Limits
Trigonometric limits involve finding the limit of functions that contain trigonometric expressions. A common challenge is to compute these limits as the variable approaches 0, which frequently requires the application of standard trigonometric limits. For example, a key concept in our exercise is the limit \(\lim_{x \to 0} \frac{1 - \cos ax}{x^2}\), which is known to be \(\frac{a^2}{2}\) by a standard trigonometric limit theorem.

Understanding these standard trigonometric limits is essential because they provide a way to simplify complex expressions to more manageable forms. This knowledge is useful in our exercise, where we used the standard limit to simplify the trigonometric component of our given function, allowing us to focus on finding the values of \(a\) and \(b\) to satisfy the given condition.
L'Hopital's Rule
In calculus, L'Hopital's rule is a popular method for evaluating the limits of indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Although L'Hopital's rule is not directly applied in our exercise, it's important to recognize situations where it could be useful. To use L'Hopital's Rule, the limit must be in the form of \(\frac{f(x)}{g(x)}\) where both \(f(x)\) and \(g(x)\) approach 0 or infinity as \(x\) approaches some value.

In our case, the term \(\frac{b}{2x^2}\) as \(x\) approaches 0 is not an indeterminate form but rather undefined because \(b\) is not a function of \(x\). Therefore, the use of L'Hopital’s rule would be inappropriate. However, knowing when and how to apply L'Hopital's rule remains a valuable tool in a mathematician's toolkit for solving a diverse range of limit problems.

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Most popular questions from this chapter

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