Problem 60 - Thomas Calculus

Question

Find the limits in Exercises \(51-66\) $$ \lim _{x \rightarrow 0^{+}}\left(1+\frac{1}{x}\right)^{x} $$

Step-by-Step Solution

Answer

The limit is \( e \).

Step 1: Identify the Indeterminate Form

The given limit is \( \lim _{x \rightarrow 0^{+}}\left(1+\frac{1}{x}\right)^{x} \). The expression \( \left(1+\frac{1}{x}\right)^{x} \) is in the form \( \left(1 + \text{something depending on } x\right)^{\text{something else depending on } x} \). When \( x \rightarrow 0^{+} \), \( \frac{1}{x} \rightarrow +\infty \), so \( \left(1+\frac{1}{x}\right)^{x} \rightarrow \left(1+\infty\right)^{0} \), which is an indeterminate form.

Step 2: Apply a Transformation to Simplify

Rewrite the expression as \( z = \left(1+\frac{1}{x}\right)^{x} \). Take the natural logarithm of both sides: \( \ln z = x \ln \left(1+\frac{1}{x}\right) \). This step allows us to transform the original expression into a form that we can evaluate.

Step 3: Expand Using a Series

For \( x \rightarrow 0^{+} \), let \( u = \frac{1}{x} \rightarrow +\infty \). This transformation allows us to use the Taylor series expansion for \( \ln(1+u) \), which is \( \ln(1+u) \approx u - \frac{u^2}{2} + \cdots \). Substituting back, \( \ln \left(1+\frac{1}{x}\right) \approx \frac{1}{x} - \frac{1}{2x^2} \).

Step 4: Substitute and Simplify

Substitute the series expansion into the transformed equation: \( \ln z = x\left(\frac{1}{x} - \frac{1}{2x^2}\right) = 1 - \frac{1}{2x} \). As \( x \rightarrow 0^{+} \), \(-\frac{1}{2x} \) goes to \( -\infty \), so \( \ln z \rightarrow 1 - \frac{1}{2}x \rightarrow 1 \).

Step 5: Evaluate the Limit

Since \( \ln z \) approaches 1, \( z = e^{\ln z} = e^{1} = e \). Thus, the original limit becomes \( \lim_{x \to 0^+} \left(1 + \frac{1}{x}\right)^x = e \).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms occur when we try to evaluate a limit and end up with an expression that does not directly provide enough information for a clear outcome. Understanding these forms is crucial for determining the limits of functions as they often appear when dealing with ratios, powers, or differences in limits.
One common example is \(1^{\infty}\), which seems confusing because raising 1 to any power usually results in 1, but when involving limits, it can approach various values.
The expression presented in the original problem, \(\left(1+\frac{1}{x}\right)^{x}\) as \(x \to 0^{+}\), leads to the indeterminate form \(\left(1+\infty\right)^{0}\).
This implies that without more analysis, we cannot determine the limit accurately just by considering the basic properties of exponents or logarithms.

Series Expansion

Series expansion is a mathematical method used to write a function as a sum of terms sourced from a sequence or series.
A key aspect in dealing with limits is utilizing power series, such as the Taylor series, which approximates functions using polynomials.

• This approach is invaluable when dealing with complex forms.
• For example, the natural logarithm function \(\ln(1+u)\) can be expanded using a Taylor series as: \(u - \frac{u^2}{2} + \cdots\).

In our problem, the function \(\ln(1+\frac{1}{x})\) is expanded as \(\frac{1}{x} - \frac{1}{2x^2}+\cdots\). This simplified form makes it easier to calculate the limit of \( \ln z\) by revealing its behavior as \(x \to 0^+\).
Such expansions are especially useful when direct calculation is challenging or impossible without simplification.

Natural Logarithms

The natural logarithm, denoted as \(\ln x\), is essential in calculus and has a base \(e\), the Euler's number.
It helps convert problems of exponentiation into multiplication, which is much simpler to handle, especially in limit problems.

• Applying \(\ln\) to both sides of an equation allows us to brush off the complicated powers.
• This step is crucial when dealing with limits involving indeterminate forms.

For the expression \(z = \left(1+\frac{1}{x}\right)^{x}\), taking the natural logarithm brings forth \(\ln z = x \ln \left(1+\frac{1}{x}\right) \), breaking down the problem to a simpler one.
From this transformation, we're able to apply series expansion, ultimately evaluating the limit without fretting over complex exponentiation directly.
Understanding how natural logs simplify exponential problems is vital in advanced calculus problems.

Exponential Function

The exponential function, often depicted as \(e^{x}\), plays a major role in calculus, providing a smooth and continuous growth curve.
It is the inverse operation of the natural logarithm and frequently appears in limit problems, especially when handling expressions involving powers.

• Converting logarithmic expressions back to exponential form can yield final results of a limit problem.
• In the original problem, after simplifying and finding \(\ln z = 1\), we apply the natural exponential function \(z = e^{1}\).

This is key, as \(e\) is a unique constant, approximately 2.718, which arises naturally in many areas of mathematics.
Mastering the exponential function enables you to effortlessly tackle limits associated with growth, decay problems, and compound interest, bridging various real-world applications to theoretical mathematics.