Problem 50 - Thomas Calculus

Question

In Exercises \(47-50,\) graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0\) .If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$ f(x)=(1+2 x)^{1 / x} $$

Step-by-Step Solution

Answer

\(f(0) = e^2\) extends the function continuously to the origin.

Step 1: Analyze the Function

We have the function \(f(x) = (1 + 2x)^{1/x}\). This function is complicated at \(x = 0\) due to the \(1/x\) power, which suggests a potential issue with continuity at this point.

Step 2: Consider Limit Approaches

To evaluate whether \(f(x)\) has a continuous extension to \(x = 0\), it is beneficial to explore the limit of \(f(x)\) as \(x\) approaches 0. Consider the limit \(\lim_{x \to 0} (1 + 2x)^{1/x}\).

Step 3: Utilize Exponential Limit Properties

The expression can be rewritten using the exponential function: \((1 + 2x)^{1/x} = e^{\frac{\ln(1+2x)}{x}}\). Now, find the limit of the exponent \( \lim_{x \to 0} \frac{\ln(1+2x)}{x} \).

Step 4: Apply L'Hopital's Rule

Because both \(\ln(1+2x)\) and \(x\) approach 0 as \(x\) approaches 0, apply L'Hopital's Rule: \(\lim_{x \to 0} \frac{\ln(1+2x)}{x} = \lim_{x \to 0} \frac{2/(1+2x)}{1} = \lim_{x \to 0} \frac{2}{1+2x} = 2\).

Step 5: Solve Logarithmic Limit

Since the limit of the exponent is found to be 2, the original limit becomes \(e^2\). Therefore, \(\lim_{x \to 0} (1+2x)^{1/x} = e^2\).

Step 6: Conclude Continous Extension

The function \(f(x)\) seems to have a continuous extension at the origin with the extended value \(f(0) = e^2\). Thus, \(f(x)\) can be made continuous everywhere by defining \(f(0) = e^2\).

Key Concepts

Limits

The concept of limits plays a fundamental role in understanding continuity and behavior of functions at specific points, especially where they may not be easily evaluated directly, like at zero in our exercise. Limits help us explore what value a function approaches as the input gets arbitrarily close to a specific point.

In the function \(f(x) = (1 + 2x)^{1/x}\), evaluating the point \(x = 0\) directly is challenging because it results in an indeterminate form \(\frac{0}{0}\). To understand what happens as \(x\) approaches zero, we use limits. We examine \(\lim_{x \to 0} (1 + 2x)^{1/x}\) to determine whether the function approaches a specific value as \(x\) nears zero.

When calculating this particular limit, we noticed that rewriting terms into exponential forms and applying mathematical strategies like L'Hopital's Rule helped resolve the indeterminate form and find that the limit approaches \(e^2\). Thus, limits enable us to reason about values and behavior of functions in crucial situations where direct substitution doesn’t work.

L'Hopital's Rule

L'Hopital's Rule is an incredibly useful tool when handling limits that result in indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It offers a method for evaluating these tricky limits by differentiating the numerator and the denominator separately, and then taking the limit again.

In our problem, we used L'Hopital's Rule to solve \(\lim_{x \to 0} \frac{\ln(1+2x)}{x}\). Initially, this is \(\frac{0}{0}\), an indeterminate form. By differentiating the numerator and denominator separately, we obtain:

• Numerator: The derivative of \(\ln(1+2x)\) is \(\frac{2}{1+2x}\).
• Denominator: The derivative of \(x\) is simply \(1\).

Then, by evaluating \(\lim_{x \to 0} \frac{2/(1+2x)}{1}\), we discovered the limit to be \(2\).

L'Hopital's Rule transformed a seemingly complex issue into something manageable, guiding the calculation towards revealing the continuous extension of the function, and showing just how impactful this rule can be when solving limits.

Exponential Functions

Exponential functions are functions of the form \(f(x) = a^{x}\), where \(a\) is a constant and \(x\) is the variable. They are fundamental in math due to their unique properties and applications in growth and decay processes.

In the context of our function \(f(x) = (1 + 2x)^{1/x}\), we transformed the expression into an exponential form to make the limit more approachable: \(e^{\frac{\ln(1+2x)}{x}}\).

This step was crucial because exponential functions allow us to express the compound interest or continuous growth idea through the constant \(e\), known as the base of natural logarithms. This method of rewriting a function as an exponential made it easier to manage the limit problem by applying L'Hopital's Rule on the exponent.

Understanding exponential functions deeply enriches our problem-solving toolkit in many areas of calculus and enriches mathematical intuition.