Problem 53 - A Graphical Approach to Precalculus with Limits

Question

Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. \(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{e^{x}-1}\)

Step-by-Step Solution

Answer

The limit exists and is equal to 2.

Step 1: Understanding the Limit

We need to find the limit of the function \( \frac{e^{2x} - 1}{e^x - 1} \) as \( x \rightarrow 0 \). This means we are examining the behavior of the function as \( x \) approaches 0 from both sides.

Step 2: Simplifying the Expression

Notice that this is an indeterminate form \( \frac{0}{0} \) at \( x = 0 \). We can use L'Hôpital's Rule to resolve the indeterminate form or proceed with algebraic manipulation. For now, let's factorize: \( e^{2x} = (e^x)^2 \) and \( e^x = 1 + x + \frac{x^2}{2} + \cdots \).

Step 3: Application of L'Hôpital's Rule

L'Hôpital's Rule states that if the limit \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) is \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \) provided this limit exists. So, differentiate the numerator and the denominator: \( f(x) = e^{2x} - 1 \), \( g(x) = e^x - 1 \).

Step 4: Differentiation

The derivative of \( f(x) = e^{2x} - 1 \) is \( f'(x) = 2e^{2x} \). The derivative of \( g(x) = e^x - 1 \) is \( g'(x) = e^x \). Substitute these into L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{2e^{2x}}{e^x} = \lim_{x \to 0} 2e^{x}. \]

Step 5: Evaluating the Limit

Now, for \( \lim_{x \to 0} 2e^x \), substitute \( x = 0 \) to get \( 2e^0 = 2 \). Thus, the limit exists and is equal to 2.

Key Concepts

Indeterminate Form

When solving limits, encountering an indeterminate form is common. The term "indeterminate form" refers to an expression that doesn't directly reveal the limit just by plugging in the value. In our problem, when substituting zero directly into the function \( \frac{e^{2x} - 1}{e^x - 1} \), both the numerator and the denominator become zero, resulting in the form \( \frac{0}{0} \).
This specific form doesn't give clear information about the limit. Various other forms, like \( \frac{\infty}{\infty} \), might also arise in limit problems and are classified as indeterminate. Indeterminate forms require extra work, such as algebraic manipulation or special techniques like L'Hôpital's Rule, to find a definitive limit.

L'Hôpital's Rule

L'Hôpital's Rule is a useful tool for resolving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) when working with limits. The rule states that if \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then it's possible to evaluate \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \) after differentiating the numerator and the denominator.
In the problem at hand, after realizing the indeterminate form, we applied L'Hôpital's Rule. We differentiated the functions as follows:

• The derivative of the numerator \( e^{2x} - 1 \) is \( 2e^{2x} \).
• The derivative of the denominator \( e^x - 1 \) is \( e^x \).

Thus, the original limit simplifies to \( \lim_{x \rightarrow 0} \frac{2e^{2x}}{e^x} = \lim_{x \rightarrow 0} 2e^x \). By substituting \( x = 0 \), the limit turns out to be \( 2 \).

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the derivative, which is essentially the rate at which a function changes. In the context of limits and L'Hôpital's Rule, differentiation helps us transform an indeterminate form into a solvable expression.
When we differentiate the functions in the original exercise, we obtain:

• For \( f(x) = e^{2x} - 1 \), the derivative is \( f'(x) = 2e^{2x} \).
• For \( g(x) = e^x - 1 \), the derivative is \( g'(x) = e^x \).

This operation allows for simplifying the complex ratio \( \frac{2e^{2x}}{e^x} \) to \( 2e^x \), which can be directly evaluated with \( x \rightarrow 0 \). Differentiation thus connects the worlds of dynamic rate changes with simplified calculations in limit problems.