Chapter 10: Problem 31
Using the Taylor series for \(f(x)=e^{x}\) around \(0,\) compute the following limit: $$\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}$$
Short Answer
Expert verified
The limit is 1.
Step by step solution
01
Write the Taylor Series Expansion
The Taylor series for \(e^x\) around \(0\) is given by the formula \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\). This series includes an infinite number of terms.
02
Substitute the Series into the Limit Expression
Replace \(e^x\) in the limit expression using the Taylor series: \[ \lim_{x \rightarrow 0} \frac{e^{x}-1}{x} = \lim_{x \rightarrow 0} \frac{(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots) - 1}{x}. \]
03
Simplify the Numerator
Simplify the expression by canceling out the 1's in the numerator: \[ \lim_{x \rightarrow 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x}. \]
04
Factor Out \(x\) from the Numerator
Factor \(x\) out of the numerator:\[ \lim_{x \rightarrow 0} \frac{x(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots)}{x}. \]
05
Cancel \(x\) in the Expression
Cancel the \(x\) in the numerator with the \(x\) in the denominator:\[ \lim_{x \rightarrow 0} (1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots). \]
06
Evaluate the Limit
Evaluate the remaining expression as \(x\) approaches 0. All terms with \(x\) in them approach 0:\[ 1 + 0 + 0 + \cdots = 1. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Evaluation
In mathematics, limit evaluation is a crucial technique for understanding the behavior of functions as they approach certain points. The idea is to find out what value a function gets closer to, as the input, often denoted as "x", becomes infinitesimally close to a particular point. This aspect of calculus enables us to determine function continuity, differentiate functions, and even compute integrals.
The limit evaluation process involves:
The limit evaluation process involves:
- Identifying the function for which you want to find the limit.
- Substituting values close to the point of interest.
- Simplifying the expression to eliminate indeterminate forms like \( \frac{0}{0} \).
- After substituting, you'll notice terms cancel out, simplifying the expression.
- Then factor out any common terms to ease simplification further.
- Finally, evaluate the remaining terms' behavior as \ x \ approaches 0.
Taylor Expansion
Taylor expansion is a powerful mathematical tool used to approximate complex functions using polynomials. This method employs derivatives of the function to build a series that approaches the desired function as more terms are added.
The Taylor series of a function \( f(x) \) about a point, say \( a \), is given by:\[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots \]It transforms functions into potentially infinite sums, making them easier to handle and solve limits, like in calculus, more straightforward.In our exercise, we used the Taylor expansion of \( e^x \) about 0:
The Taylor series of a function \( f(x) \) about a point, say \( a \), is given by:\[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots \]It transforms functions into potentially infinite sums, making them easier to handle and solve limits, like in calculus, more straightforward.In our exercise, we used the Taylor expansion of \( e^x \) about 0:
- This series is \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \).
- By substituting this into our limit expression, we simplified the numerator into an algebraic form that allowed for easy limit evaluation.
- This simplification demonstrates how Taylor expansions can turn complex functions into simpler polynomial forms that are much easier to work with.
Exponential Function
The exponential function, denoted as \( e^x \), is a fundamental mathematical function known for its unique properties and importance in various fields like calculus, complex analysis, and number theory.
Some key features include:
Some key features include:
- Its value is always positive, stretching from zero to infinity.
- The rate of growth is proportional to its current value, which is why it's seen in nature and economics.
- It has a remarkable property where its derivative is the function itself, i.e., \( \frac{d}{dx}(e^x) = e^x \).