Chapter 11: Problem 26
Use power series rather than I'Hôpital's rule to evaluate the given limit. \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\arctan x}\)
Short Answer
Expert verified
The limit is 1.
Step by step solution
01
Expand the Series for Exponential Functions
First, expand the exponential functions using their power series around \( x = 0 \): \[e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \]\[e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots\]Now, subtract these two expansions:\[e^{x} - e^{-x} = (1 + x + \frac{x^2}{2} + \frac{x^3}{6}) - (1 - x + \frac{x^2}{2} - \frac{x^3}{6}) = 2x + \frac{x^3}{3} + \cdots\]
02
Simplify the Numerator
Given the expression from Step 1, we have the numerator:\[e^{x} - e^{-x} - 2x = (2x + \frac{x^3}{3} + \cdots) - 2x = \frac{x^3}{3} + \cdots\]This shows that terms after subtracting \(2x\) vanish up to the order \(x^3\).
03
Expand the Series for Arctan Function
The power series expansion for \(\arctan x\) around \(x=0\) is:\[arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\]The denominator \(x - \arctan x\) becomes:\[x - (x - \frac{x^3}{3} + \cdots) = \frac{x^3}{3} - \cdots\]
04
Evaluate the Limit Using the Series
Now we evaluate the limit using the simplified series:\[\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x- ext{arctan} x} = \lim _{x \rightarrow 0} \frac{\frac{x^3}{3} + \cdots}{\frac{x^3}{3} - \cdots}\]Given that both numerator and denominator's leading term is \(\frac{x^3}{3}\), the fractional limit as \(x \to 0\) simplifies to:\[\frac{\frac{x^3}{3}}{\frac{x^3}{3}} = 1\]
05
Conclusion
The limit evaluates to given the power series expansions and simplifications, thus:\[\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x- ext{arctan} x} = 1\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Functions
Exponential functions are a fundamental class of functions in calculus, characterized by the constant base raised to the power of a variable exponent. The function \( e^x \), where \( e \) is Euler's number (approximately 2.71828), is particularly special because its derivative and integral are the same function. This property makes \( e^x \) highly useful in calculus and mathematical modeling.
When considering power series expansions, \( e^x \) can be expressed as an infinite series:
When considering power series expansions, \( e^x \) can be expressed as an infinite series:
- \( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \)
- \( e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots \)
Arctan Function
The arctan function, short for arctangent, is the inverse of the tangent function. It is used to determine the angle whose tangent is a given number, producing values between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). It is prevalent in trigonometry and calculus due to its role in solving trigonometric equations and evaluating limits.
Expressed as a power series, the arctan function can be expanded for small values of \( x \):
Expressed as a power series, the arctan function can be expanded for small values of \( x \):
- \( \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \)
Limit Evaluation
Evaluating limits is a central concept in calculus, often applied to determine the behavior of functions as they approach particular points. When direct substitution in a function results in an indeterminate form, such as \( \frac{0}{0} \), alternative techniques like series expansion or L'Hôpital's Rule are employed.
In the context of this exercise, we utilize power series to manage the indeterminate form:
In the context of this exercise, we utilize power series to manage the indeterminate form:
- Rewriting both the numerator and denominator using their series.
- Simplifying these series to evaluate the limit as \( x \to 0 \).
Series Expansion
A series expansion expresses a function as a sum of terms calculated from its derivatives at a specific point. This transformation allows mathematicians to approximate functions that are otherwise difficult to compute directly.
For functions like \( e^x \) and \( \arctan x \), series expansions are used to express them as power series:
For functions like \( e^x \) and \( \arctan x \), series expansions are used to express them as power series:
- They provide a polynomial approximation of a function, focusing on the behavior of functions near a point, commonly \( x = 0 \).
- They are a vital tool in calculus for solving limits, integrating complex functions, and modeling real-world phenomena.