Problem 30 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

\(\frac{\ln(3)}{\ln(2)}\)

Step 1: Check Indeterminate Form

First, we substitute \(x = 0\) into the given limit expression \(\frac{3^x - 1}{2^x - 1}\). This gives us \(\frac{3^0 - 1}{2^0 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}\), which is an indeterminate form. This means we can apply l'Hôpital's rule.

Step 2: Differentiate the Numerator and Denominator

According to l'Hôpital's rule, we differentiate the numerator and the denominator separately. The derivative of the numerator \(3^x - 1\) is \(\ln(3) \cdot 3^x\). The derivative of the denominator \(2^x - 1\) is \(\ln(2) \cdot 2^x\).

Step 3: Apply l'Hôpital's Rule and Compute the Limit

Substitute the derivatives back into the limit expression to get \(\lim_{x \rightarrow 0} \frac{\ln(3) \cdot 3^x}{\ln(2) \cdot 2^x}\). Simplify the expression: \(\frac{\ln(3)}{\ln(2)} \cdot \frac{3^x}{2^x}\). As \(x\) approaches \(0\), \(\frac{3^x}{2^x} \rightarrow \frac{3^0}{2^0} = 1\). Thus, the limit becomes \(\frac{\ln(3)}{\ln(2)}\).

Step 4: Final Solution

The value of the original limit is \(\frac{\ln(3)}{\ln(2)}\). Thus, \(\lim_{x \rightarrow 0} \frac{3^x - 1}{2^x - 1} = \frac{\ln(3)}{\ln(2)}\).

Key Concepts

Limits

In mathematics, a limit describes the value that a function approaches as the input approaches some value. When dealing with limits, it's like asking where a roller coaster is heading as it zooms down the track, without worrying about whether it actually stops there. For the expression given in the exercise, \[\lim_{x \to 0} \frac{3^x - 1}{2^x - 1},\] the limit helps us understand what happens to the fraction as \(x\) gets very close to zero.

• To find a limit, substitute the approaching value into the function.
• If a substitution leads to a numerical answer, that's your limit.
• Sometimes, like in this problem, substitution leads to an indeterminate form, such as \(\frac{0}{0}\).

This is where special techniques come in handy, such as l'Hôpital's Rule, to resolve what the limit truly is.

Indeterminate Forms

Indeterminate forms arise when directly substituting the limiting value into the function doesn't yield a meaningful result. Common forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and more. In this example, plugging \(x = 0\) into the fraction \(\frac{3^x - 1}{2^x - 1}\) gives us \(\frac{0}{0}\), an indeterminate form.

• When you encounter an indeterminate form, the answer isn't obvious.
• You need additional methods to find the limit, like algebraic manipulation or l'Hôpital's Rule.
• Indeterminate forms can occur in various mathematical contexts, not just in limits.

Understanding and identifying these forms is crucial since they signal the need for more sophisticated techniques to unravel the true behavior of functions near specific points.

Differentiation

Differentiation is a fundamental concept in calculus which involves finding the derivative of a function. The derivative represents an instantaneous rate of change, similar to finding how fast a car is moving at an exact moment.For limits approaching indeterminate forms like \(\frac{0}{0}\), differentiation becomes essential.

• The core idea behind l'Hôpital's Rule is differentiating the numerator and denominator separately when you encounter \(\frac{0}{0}\).
• In our example, differentiating \(3^x - 1\) gives \(\ln(3) \cdot 3^x\), and \(2^x - 1\) differentiates to \(\ln(2) \cdot 2^x\).
• Once each part is differentiated, you recompute the limit using these derivatives instead of the original functions.

This technique often simplifies the problem, allowing you to determine a precise limit even when direct substitution fails. Differentiation thus acts as a powerful tool in solving complex expressions where simple calculations aren't enough.