Problem 31 - Calculus

Question

Using the Taylor series for \(f(x)=e^{x}\) around \(0,\) compute the following limit: $$\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: 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.

Step 2: Substitute the Series into the Limit Expression

Replace \(e^x\) in the limit expression using the Taylor series: \[ \lim_{x \rightarrow 0} \frac{e^{x}-1}{x} = \lim_{x \rightarrow 0} \frac{(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots) - 1}{x}. \]

Step 3: Simplify the Numerator

Simplify the expression by canceling out the 1's in the numerator: \[ \lim_{x \rightarrow 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x}. \]

Step 4: Factor Out \(x\) from the Numerator

Factor \(x\) out of the numerator:\[ \lim_{x \rightarrow 0} \frac{x(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots)}{x}. \]

Step 5: Cancel \(x\) in the Expression

Cancel the \(x\) in the numerator with the \(x\) in the denominator:\[ \lim_{x \rightarrow 0} (1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots). \]

Step 6: Evaluate the Limit

Evaluate the remaining expression as \(x\) approaches 0. All terms with \(x\) in them approach 0:\[ 1 + 0 + 0 + \cdots = 1. \]

Key Concepts

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:

• 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 \( \frac{0}{0} \).

For example, consider the limit \( \lim_{x \rightarrow 0} \frac{e^{x}-1}{x} \). To evaluate this, we substitute the expression \( e^x - 1 \) with its Taylor series expansion:

• 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.

In this instance, the solution showed that all additional terms become negligible, resulting in a clear evaluation of the limit as 1.

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:

• 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.

Using Taylor expansion helps visualize how functions behave, especially near particular points, aiding in solving crucial mathematical problems.

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:

• 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., \( \frac{d}{dx}(e^x) = e^x \).

Because of these properties, \( e^x \) shows up in a wide range of applications, like population growth models, compound interest calculations, and even in solving differential equations.In our limit evaluation, the exponential function's Taylor series provided an alternative view of the function, especially near 0, showing how quickly it simplifies to a foundational concept like a limit result of 1. Understanding \( e^x \) equips students to tackle various problems in calculus and beyond.