Chapter 9: Problem 42
Find a power series expansion for \(\left(e^{x}-1-x\right) / x^{2}\) and use it to evaluate $$ \lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}} $$
Short Answer
Expert verified
The limit is \(\frac{1}{2}\).
Step by step solution
01
Expand the Exponential Function
First, we need to find the Taylor series expansion for the exponential function, which is given by: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots \]
02
Compute \\(e^x - 1 - x\\)
Using the expansion from Step 1, compute \(e^x - 1 - x\):\[ e^x - 1 - x = \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \right) - 1 - x \]Cancelling out the 1 and x terms, we get:\[ \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \ldots \]
03
Divide by \\(x^2\\)
Now, divide the expression from Step 2 by \(x^2\) to find the power series for \((e^x - 1 - x)/x^2\):\[ \frac{1}{x^2} \cdot \left( \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \ldots \right) \]This simplifies to:\[ \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} + \ldots \]
04
Evaluate the Limit
To find \(\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\), evaluate the limit of the power series as \(x\) approaches 0. Since the limit of a power series as \(x \to 0\) is influenced only by the constant term:\[ \lim_{x \to 0} \left( \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} + \ldots \right) = \frac{1}{2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series Expansion
Power series expansion is a method used in mathematics to express functions as an infinite sum of terms calculated from the function's derivatives at a single point. It's like breaking down a complex function into simpler building blocks that we can easily work with. A power series for a function can be represented as:
- \( f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots \)
Taylor Series
A Taylor series is a specific kind of power series expansion of a function about a point. It approximates functions with a polynomial that incorporates information from the function’s derivatives at a single point. The formula for the Taylor series of a function \(f(x)\) centered at \(a\) is:
- \(f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \)
Exponential Function Expansion
Exponential function expansion involves expressing the function \(e^x\) as an infinite series of terms, making it much easier to use in a variety of calculus problems. The exponential function \(e^x\) has a well-known expansion derived from the Taylor series, which looks like:
- \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots\)
Calculus Limits
Calculus limits are a fundamental concept that describe the behavior of a function as it approaches a particular point. Limits help us understand the tendencies of a function's value as the input approaches some number. In our current exercise, the limit we are evaluating is:
- \(\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\)