Chapter 20: Problem 10
If the function \(f\) defined as \(f(x)=\frac{1}{x}-\frac{k-1}{e^{2 x}-1}\) \(x \neq 0\), is continuous at \(x=0\), then the ordered pair \((k, f(0))\) is equal to? \(\begin{array}{llll}\text { (a) }(3,1) & \text { (b) }(3,2) & \text { (c) }\left(\frac{1}{3}, 2\right) & \text { (d) }(2,1)\end{array}\)
Short Answer
Step by step solution
Explore Continuity Condition
Analyze Individual Terms
Simplify the Exponential Term
Combine Terms and Find Limit
Solve for \(k\)
Determine \(f(0)\) Value
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits in Calculus
Consider the function \(f(x) = \frac{1}{x} - \frac{k-1}{e^{2x} - 1}\). To examine its continuity at \(x = 0\), we evaluate the limit of \(f(x)\) as \(x\) approaches 0. The critical aspect is ensuring that
- The limit \(\lim_{{x \to 0}} f(x)\) exists.
- The limit equals \(f(0)\).
Exponential Functions
In the exercise, the exponential function \(e^{2x}\) plays a pivotal role. As we expand \(e^{2x}\) near zero using series expansion, we approximate it by its Taylor series:
- Near zero, \(e^{2x} \approx 1 + 2x + \frac{(2x)^2}{2} + \ldots\).
- For very small values of \(x\), higher-order terms become negligible.
- This ultimately simplifies our problem by allowing us to substitute and simplify \(e^{2x} - 1\) as approximately \(2x\).
Series Expansion
The series expansion of an exponential function, like \(e^{2x}\), helps by providing:
- A way to approximate the behavior of functions close to a point.
- A means to transform complex expressions into simpler, polynomial-like forms.
Algebraic Manipulation
In this exercise, by substituting the approximation \(e^{2x} - 1 \approx 2x\), the problem simplifies to evaluating:
- \(f(x) \approx \frac{1}{x} - \frac{k-1}{2x} = \frac{3-k}{2x}\)
This example demonstrates how algebra can streamline evaluating limits and finding the continuity of a function. It stresses the importance of simplifying expressions to reach a solution efficiently and accurately.