Problem 30 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow 0} \frac{3^{x}-1}{2^{x}-1}$$

Step-by-Step Solution

Answer

The limit is \(\frac{\ln(3)}{\ln(2)}\)."}

Step 1: Understand the Problem

We need to find the limit as \(x\) approaches \(0\) for the expression \(\frac{3^{x}-1}{2^{x}-1}\). At \(x=0\), this expression has the form \(\frac{0}{0}\), which allows us to apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if a limit is in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\), provided this new limit exists. We will differentiate the numerator and denominator separately.

Step 3: Differentiate the Numerator and Denominator

Differentiate the numerator \(f(x) = 3^x - 1\) with respect to \(x\). The derivative is \(f'(x) = 3^x \ln(3)\). Differentiate the denominator \(g(x) = 2^x - 1\) with respect to \(x\). The derivative is \(g'(x) = 2^x \ln(2)\).

Step 4: Evaluate the Limit of the Derivatives

Now find the limit of the derivatives: \( \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{3^x \ln(3)}{2^x \ln(2)} \). At \(x=0\), this becomes \(\frac{1 \cdot \ln(3)}{1 \cdot \ln(2)} = \frac{\ln(3)}{\ln(2)}\).

Step 5: Conclude with the Result

The limit \(\lim_{x \to 0} \frac{3^{x} - 1}{2^{x} - 1}\) evaluates to \(\frac{\ln(3)}{\ln(2)}\) when applying L'Hôpital's Rule.

Key Concepts

Limit Evaluation

When we're faced with evaluating a limit, we often start by substituting the approaching value directly into our expression. However, sometimes this gives us an indeterminate form, such as \(\frac{0}{0}\). That's where calculus tools like L'Hôpital's Rule become handy. To use this rule, confirm that your expression is indeed in an indeterminate form. Begin by substitifying the limit point into both the numerator and the denominator. If both result in 0, L'Hôpital's Rule can be applied.

• Always re-check your expression after using L'Hôpital's Rule to make sure it no longer remains indeterminate.
• Not all limits are 0/0 or ∞/∞, so L'Hôpital's Rule isn't always applicable.

Applying the rule means differentiating the numerator and denominator separately. Evaluate the new limit with these derivatives. This process might need repeating if the expression remains indeterminate.

Derivatives

Derivatives are the backbone of calculus, and they help describe how functions change. When using L'Hôpital's Rule, we rely heavily on this concept since it involves finding the derivatives of the numerator and the denominator separately. Consider exponential functions like \(3^x\) and \(2^x\), their derivatives involve the original function multiplied by the natural logarithm of their bases.

• The derivative of \(3^x\) is \(3^x \ln(3)\).
• The derivative of \(2^x\) is \(2^x \ln(2)\).

Make sure to differentiate correctly, as this is crucial for finding the correct limit using L'Hôpital's Rule. The accuracy of your derivatives directly impacts the correctness of the final limit evaluation.

Exponential Functions

Exponential functions have the general form \(a^x\), where \(a\) is a constant. These functions grow very quickly; faster than polynomial functions. When you're dealing with limits involving exponential functions, understanding their properties can help simplify problems.Each exponential function has a unique growth rate, often expressed using the natural logarithm. This is crucial when determining derivatives because:

• The derivative involves product of the function itself and the natural logarithm of the base \(a\), expressed as \(a^x \ln(a)\).
• This characteristic allows exponential functions to have their unique exponential acceleration behavior less impacted by multiplication adjustments in derivatives.

In limit problems, exponential functions can cause indeterminate forms that require careful handling using logarithmic properties and derivative calculations, just as we've seen with L'Hôpital's Rule.