Problem 84 - Thomas Calculus in SI Units

Question

This exercise explores the difference between the limit $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x^{2}}\right)^{x}$$ and the limit $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$$ a. Use l'Hôpital's Rule to show that $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$$ b. Graph $$f(x)=\left(1+\frac{1}{x^{2}}\right)^{x} \text { and } g(x)=\left(1+\frac{1}{x}\right)^{x}$$ together for \(x \geq 0 .\) How does the behavior of \(f\) compare with that of \(g\) ? Estimate the value of \(\lim _{x \rightarrow \infty} f(x)\) c. Confirm your estimate of \(\lim _{x \rightarrow \infty} f(x)\) by calculating it with I'Hôpital's Rule.

Step-by-Step Solution

Answer

\(\lim_{x \to \infty} (1+\frac{1}{x^2})^x = 1\).

Step 1: Limit Evaluation with L'Hôpital's Rule for \(g(x)\)

First, we need to evaluate \(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e\) using the exponential limit definition. Let \(y = \left(1+\frac{1}{x}\right)^{x}\), and take the natural logarithm of both sides to get \(\ln y = x \ln\left(1 + \frac{1}{x}\right)\).As \(x \to \infty\), the expression \(\ln\left(1 + \frac{1}{x}\right)\) can be rewritten using the approximation \(\ln(1+u) \approx u\) when \(u\) is small. Thus, \(x \ln\left(1+\frac{1}{x}\right) \approx \frac{x}{x} = 1\).Apply L'Hôpital's Rule to \(x \ln\left(1 + \frac{1}{x}\right) = \frac{\ln(1 + \frac{1}{x})}{1/x}\) because both the numerator and the denominator go to zero as \(x \to \infty\). Take the derivative of the numerator and the denominator:- Derivative of \(\ln(1 + \frac{1}{x})\) with respect to \(x\) is \(-\frac{1}{x(x + 1)}\).- Derivative of \(\frac{1}{x}\) with respect to \(x\) is \(-\frac{1}{x^2}\).Applying L'Hôpital's Rule, \[\lim_{x \to \infty} \frac{\ln(1 + \frac{1}{x})}{1/x} = \lim_{x \to \infty} \frac{-\frac{1}{x(x + 1)}}{-\frac{1}{x^2}} = \lim_{x \to \infty} \frac{x^2}{x(x+1)} = \lim_{x \to \infty} \frac{x}{x+1} = 1.\]Thus, \( \ln y = 1 \) means \( y = e \). Then, \( \lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e \).

Step 2: Graphical Comparison of \(f(x)\) and \(g(x)\)

To compare the behavior of \(f(x) = \left(1+\frac{1}{x^2}\right)^x\) and \(g(x) = \left(1+\frac{1}{x}\right)^x\), plot both functions for \(x \geq 0\).- The graph of \(g(x)\) should show a convergence to \(e\) as \(x\) increases.- The graph of \(f(x)\), however, tends toward a number slightly greater than 1 but not close to \(e\).Observation of the graph should suggest that as \(x\) approaches \(\infty\), \(f(x)\) doesn't approach \(e\) and appears to converge to 1 or a number slightly greater than 1.

Step 3: Calculating \(\lim _{x \rightarrow \infty} f(x)\) with L'Hôpital's Rule

To confirm our estimation of \(\lim _{x \rightarrow \infty} f(x)\), let \(y = \left(1+\frac{1}{x^2}\right)^{x}\).Take the natural logarithm: \(\ln y = x \ln\left(1 + \frac{1}{x^2}\right)\). This becomes an indeterminate form as \(x \to \infty\). Rewrite it as \(\frac{\ln(1 + \frac{1}{x^2})}{1/x}\).Apply L'Hôpital's Rule:- The derivative of \(\ln(1 + \frac{1}{x^2})\) with respect to \(x\) is \(-\frac{2}{x^3(1 + \frac{1}{x^2})}\).- The derivative of \(\frac{1}{x}\) with respect to \(x\) is \(-\frac{1}{x^2}\).Applying L'Hôpital's rule,\[\lim_{x \to \infty} \frac{\ln(1 + \frac{1}{x^2})}{1/x} = \lim_{x \to \infty} \frac{-\frac{2}{x^3(1 + \frac{1}{x^2})}}{-\frac{1}{x^2}} = \lim_{x \to \infty} \frac{2x^2}{x^3(1+\frac{1}{x^2})} = \lim_{x \to \infty} \frac{2}{x(1+\frac{1}{x^2})} = 0.\]Hence, \(\ln y = 0\) implies \(y = e^0 = 1\).So \(\lim _{x \to \infty} f(x) = 1\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that initially appear as indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter these forms, L'Hôpital's Rule allows you to take the derivative of the numerator and the derivative of the denominator separately, which often simplifies the limit to a form that can be directly evaluated.
Sometimes, repeatedly applying L'Hôpital's Rule is necessary until a determinate form emerges.
This technique works under specific conditions:

• The original functions must be differentiable.
• The new limit after taking derivatives exists.

This method was originally used in the solution to handle the indeterminate forms emerging in both functions, \( g(x) \) and \( f(x) \), by taking the appropriate derivatives.

Exponential Functions

Exponential functions, like the ones in our exercise, often involve expressions raised to a limit, such as \( \left(1 + \frac{1}{x}\right)^x \) or \( \left(1 + \frac{1}{x^2}\right)^x \). These expressions illustrate how a tiny addition raised to a large power can yield significant outcomes, which is the core of many exponential growth scenarios.
The mathematical constant \( e \) is crucial here, frequently appearing in limits involving expressions of the form \( \left(1 + \frac{1}{n}\right)^n \) as \( n \to \infty \). These functions model phenomena where the quantity grows proportionally to its size, leading to concepts such as continuous growth.
Understanding exponential functions is vital for evaluating limits analytically and graphically, which is a key part of solving problems that depict real-world continuous changes.

Indeterminate Forms

Indeterminate forms are expressions that don't initially provide a clear path to evaluate limits, often appearing as \( 0/0 \), \( \infty/\infty \), or others like \( \infty - \infty \). These forms require special techniques, like algebraic manipulation or L'Hôpital's Rule, to resolve.
In the context of the exercise, the limits of the functions involved can result in indeterminate expressions.
By carefully manipulating these forms with tools like L'Hôpital's Rule, we can transform them into determinate forms that reveal the actual limit.
Simplifying these forms can be tricky, requiring steps like rationalization or logarithmic transformation to assist in the application of derivatives.
Recognizing indeterminate forms is critical, as it sets the stage for choosing the correct method to find the limit.

Graphical Analysis

Graphical analysis offers a visual perspective on complex mathematical problems, making abstract concepts more intuitive. Graphing functions like \( f(x) = \left(1 + \frac{1}{x^2}\right)^x \) and \( g(x) = \left(1 + \frac{1}{x}\right)^x \) help us understand and compare their behavior as \( x \to \infty \).
By plotting these functions:

• \( g(x) \) visibly converges to \( e \), reinforcing our analytical finding.
• \( f(x) \) grows towards a number slightly greater than 1 but less than \( e \), illustrating its differing limit.

Such visualizations also provide insights into a function's concavity, rate of growth, and other characteristics.
Understanding these graphical behaviors can be a valuable check against errors in analytical work, helping assure that the algebraic manipulations align with the graph's predictions.