Problem 12 - Essential Calculus Early Transcendentals

Question

\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{t \rightarrow 0} \frac{8^{t}-5^{t}}{t}$$

Step-by-Step Solution

Answer

The limit is \(\ln\left(\frac{8}{5}\right)\).

Step 1: Identify the Indeterminate Form

First, substitute \(t = 0\) into the expression \(\frac{8^t - 5^t}{t}\). We find that \(8^0 = 1\) and \(5^0 = 1\), so the expression becomes \(\frac{1 - 1}{0} = \frac{0}{0}\). This is an indeterminate form, which indicates potential use of l'Hospital's Rule.

Step 2: Check Conditions for l'Hospital's Rule

Verify the conditions required for applying l'Hospital's Rule. Both the numerator and the denominator are differentiable and approach 0 as \(t \rightarrow 0\). Thus, we can apply l'Hospital's Rule.

Step 3: Differentiate Numerator and Denominator

Differentiate the numerator \(8^t - 5^t\) using the chain rule. The derivative of \(8^t\) is \(8^t \ln(8)\), and the derivative of \(5^t\) is \(5^t \ln(5)\). The derivative of the denominator \(t\) is 1.So, the derivative of the fraction becomes \(\frac{8^t \ln(8) - 5^t \ln(5)}{1}\).

Step 4: Evaluate the New Limit

Substitute \(t = 0\) into the derived expression. \[\lim_{t \rightarrow 0} \frac{8^t \ln(8) - 5^t \ln(5)}{1} = 8^0 \ln(8) - 5^0 \ln(5) = 1 \cdot \ln(8) - 1 \cdot \ln(5) = \ln(8) - \ln(5)\].

Step 5: Simplify the Expression

Use the logarithmic identity \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\) to simplify the final result:\(\ln(8) - \ln(5) = \ln\left(\frac{8}{5}\right)\). So, the limit is \(\ln\left(\frac{8}{5}\right)\).

Key Concepts

Limits in Calculus

Limits are a foundational concept in Calculus, playing a crucial role in defining derivatives and integrals. The limit of a function describes its behavior as the input approaches a certain value. This approach helps us understand the function's behavior near points of interest when direct computation becomes challenging or impossible.

For the problem at hand, we focus on the limit \[\lim_{t \rightarrow 0} \frac{8^t - 5^t}{t}\]When dealing with limits, particularly those involving potential indeterminate forms (like \( \frac{0}{0}\)), it's common to explore algebraic simplification, series expansion, or the application of rules like l'Hospital's Rule.

Understanding how a function approaches infinity or how values evolve infinitesimally near a point equips you with powerful tools to gauge and predict function behavior. In this exercise, recognizing that substituting \( t = 0 \) yields \( \frac{0}{0} \) is crucial, prompting us to explore further with more advanced techniques like l'Hospital's Rule.

Indeterminate Forms

Indeterminate forms arise in calculus when direct substitution in a limit leads to ambiguous expressions, such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), or \(\infty - \infty\). These expressions don't directly point to a clear numerical limit, which necessitates more sophisticated methods to find the actual limit value.

The given exercise results in the indeterminate form \(\frac{0}{0}\) upon substituting \(t = 0\) into \(\frac{8^t - 5^t}{t}\). Recognizing these forms allows us to consider the use of l'Hospital's Rule. This rule is a powerful tool that leverages differentiation to resolve indeterminate forms in limits.

It’s important to check the conditions before applying l'Hospital's Rule:

• Both the numerator and denominator first approach 0 or infinity.
• The functions involved must be differentiable in a neighborhood around the point.

In this problem, both conditions are met, allowing us to differentiate and resolve the indeterminate form.

Logarithms in Calculus

Logarithms are pervasive in calculus, offering methods to simplify complex expressions and solve exponential equations. In the context of limits and derivatives, they frequently appear in the differentiation of exponential functions due to their unique properties.

During the derivative process in the exercise, we differentiate expressions like \(8^t - 5^t\), which introduce logarithms via the chain rule. Specifically, the derivative of an exponential function \(a^t\) is \(a^t \ln(a)\). Thus, we derive:

• The derivative of \(8^t\) becomes \(8^t \ln(8)\).
• The derivative of \(5^t\) becomes \(5^t \ln(5)\).

These logarithmic terms provide a clearer path to evaluate limits involving exponential functions.

Once we evaluate the derived limit, we calculate\[\lim_{t \rightarrow 0} \left(8^t \ln(8) - 5^t \ln(5)\right)\]At \(t = 0\), this reduces to \(\ln(8) - \ln(5)\), leading to the simplified result using the property \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\), giving us \(\ln\left(\frac{8}{5}\right)\). This property of logarithms showcases their usefulness in simplifying complex calculations across various calculus problems.