Problem 63 - Single Variable Calculus

Question

Find the limit of the sequence \(\left\\{\left(1+\frac{2}{n}\right)^{3 n}\right\\}\). Confirm your results visually by plotting the graph of $$ f(x)=\left(1+\frac{2}{x}\right)^{3 x} $$

Step-by-Step Solution

Answer

In summary, the limit of the sequence $\left\\{\left(1+\frac{2}{n}\right)^{3n}\right\\}$ is $e^6$. This result is confirmed visually by plotting the graph of the function $f(x)=\left(1+\frac{2}{x}\right)^{3x}$ and observing that it converges to $e^6$ as $x$ approaches infinity.

Step 1: Rewrite the function as an exponential function with base \(e\)

Rewrite the function \(f(x)\) using the property that \(a^b = e^{\ln a^b}\). Applying this property, we have: $$ f(x) = \left(1+\frac{2}{x}\right)^{3x} = e^{\ln\left(\left(1+\frac{2}{x}\right)^{3x}\right)} $$ Now, use the property \(\ln a^b = b\ln a\). This gives: $$ f(x) = e^{3x\ln\left(1+\frac{2}{x}\right)} $$

Step 2: Apply the limit

Consider the limit as \(x \rightarrow \infty\): $$ \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} e^{3x\ln\left(1+\frac{2}{x}\right)} $$ Then, focus on the exponent of the above expression: $$ \lim_{x \rightarrow \infty} 3x\ln\left(1+\frac{2}{x}\right) $$

Step 3: Use L'Hôpital's Rule

To find the limit in Step 2, we can use L'Hôpital's Rule by taking the derivatives of both the numerator and denominator with respect to \(x\). First, rewrite the limit as a quotient: $$ \lim_{x \rightarrow \infty} \frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{3x}} $$ Now, take the derivatives of the numerator and denominator: $$ \frac{d}{dx}\left(\ln\left(1+\frac{2}{x}\right)\right) = \frac{-2}{x\left(x+2\right)} $$ $$ \frac{d}{dx}\left(\frac{1}{3x}\right) = -\frac{1}{3x^2} $$ Then, apply L'Hôpital's Rule to the limit: $$ \lim_{x \rightarrow \infty} \frac{\frac{-2}{x\left(x+2\right)}}{-\frac{1}{3x^2}} = \lim_{x \rightarrow \infty} \frac{6}{x+2} = 6 $$

Step 4: Find the limit of the function

Now, substitute the result from Step 3 back into the expression for \(\lim_{x \rightarrow \infty} f(x)\): $$ \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} e^{3x\ln\left(1+\frac{2}{x}\right)} = e^6 $$ The limit of the sequence is: $$ \lim_{n \rightarrow \infty} \left(1+\frac{2}{n}\right)^{3n} = e^6 $$

Step 5: Plot the graph of the function

Next, plot the graph of the function \(f(x) = \left(1+\frac{2}{x}\right)^{3x}\) to confirm the result visually. When you plot function for large values of \(x\), you will see that the graph converges to \(e^6\) as \(x\) approaches infinity, which agrees with our result from Step 4. The function indeed approaches \(e^6\) and our solution is confirmed visually.

Key Concepts

Exponential functions

Exponential functions are a type of mathematical expression where a constant base is raised to a variable exponent, usually written in the form \( a^x \). These functions are known for their rapid growth or decay, depending on the base value.

In the given exercise, the sequence \( \left\{\left(1+\frac{2}{n}\right)^{3n}\right\} \) is transformed into an exponential form involving the base \( e \), the natural exponential function. This transformation is done to facilitate the application of limits and differentiation, which are tools frequently used to analyze such problems.

By rewriting the function as \( e^{3x\ln\left(1+\frac{2}{x}\right)} \), we leverage the property \( a^b = e^{b\ln a} \), making it easier to take the limit as \( x \to \infty \).

• This conversion highlights the power of exponential functions in calculus, allowing complex expressions to be handled more simply.
• It also underscores the importance of logarithmic functions in expressing growth and decay.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits of indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In the exercise, this rule is used to handle an indeterminate form that arises in the limit process.

We encounter this when evaluating \( \lim_{x \rightarrow \infty} \frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{3x}} \). Initially, both the numerator and denominator approach zero, creating an \( \frac{0}{0} \) form, which is ideal for applying L'Hôpital's Rule.

• The derivatives are taken: \( \frac{d}{dx}\ln\left(1+\frac{2}{x}\right) = \frac{-2}{x(x+2)} \) and \( \frac{d}{dx}\left(\frac{1}{3x}\right) = -\frac{1}{3x^2} \).
• Once these derivatives are calculated, the limit is explored again, yielding \( \frac{6}{x+2} \) as \( x \) approaches infinity. This resolves to the limit being 6.

This rule simplifies the process of calculating such limits by converting the problem into something more manageable—using differentiation instead of directly attempting the limit analysis.

Graphical confirmation

Plotting graphs of mathematical functions provides a visual way to confirm analytical results. Visual verification is particularly useful in understanding the behavior of sequences and functions at boundaries, such as limits approaching infinity.

In our scenario, the function \( f(x) = \left(1+\frac{2}{x}\right)^{3x} \) is graphed to observe its behavior as \( x \to \infty \).

• By plotting, we can see how the function behaves and progressively approaches the predicted limit value of \( e^6 \).
• Graphical tools help in detecting any deviations or errors in analytical calculations.

The graph confirms the analytical solution, showing that as \( x \) becomes very large, the function indeed approaches \( e^6 \). This graphical component is crucial not only for confirmation but also for aiding intuitive understanding.