Problem 21 - Calculus: Early Transcendentals

Question

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places). \( \displaystyle \lim_{t \to 0}\frac{e^{5t} - 1}{t} \), \( t = \pm 0.5, \pm 0.1, \pm 0.01, \pm 0.001, \pm 0.0001 \)

Step-by-Step Solution

Answer

The limit is approximately 5.

Step 1: Understanding the Function

We need to evaluate the limit \( \lim_{t \to 0}\frac{e^{5t} - 1}{t} \). The given function is \( f(t) = \frac{e^{5t} - 1}{t} \), and we will evaluate this function at specific values of \( t \) approaching 0 from both positive and negative sides.

Step 2: Evaluate the Function at \( t = 0.5 \) and \( t = -0.5 \)

Calculate \( \frac{e^{5(0.5)} - 1}{0.5} \) and \( \frac{e^{5(-0.5)} - 1}{-0.5} \).Evaluating gives approximately: - For \( t = 0.5 \), value is 29.09802.- For \( t = -0.5 \), value is 21.48682.

Step 3: Evaluate the Function at \( t = 0.1 \) and \( t = -0.1 \)

Calculate \( \frac{e^{5(0.1)} - 1}{0.1} \) and \( \frac{e^{5(-0.1)} - 1}{-0.1} \).Evaluating gives approximately: - For \( t = 0.1 \), value is 6.64872.- For \( t = -0.1 \), value is 4.87706.

Step 4: Evaluate the Function at \( t = 0.01 \) and \( t = -0.01 \)

Calculate \( \frac{e^{5(0.01)} - 1}{0.01} \) and \( \frac{e^{5(-0.01)} - 1}{-0.01} \).Evaluating gives approximately: - For \( t = 0.01 \), value is 5.12711.- For \( t = -0.01 \), value is 4.87823.

Step 5: Evaluate the Function at \( t = 0.001 \) and \( t = -0.001 \)

Calculate \( \frac{e^{5(0.001)} - 1}{0.001} \) and \( \frac{e^{5(-0.001)} - 1}{-0.001} \).Evaluating gives approximately: - For \( t = 0.001 \), value is 5.00501.- For \( t = -0.001 \), value is 4.99500.

Step 6: Evaluate the Function at \( t = 0.0001 \) and \( t = -0.0001 \)

Calculate \( \frac{e^{5(0.0001)} - 1}{0.0001} \) and \( \frac{e^{5(-0.0001)} - 1}{-0.0001} \).Evaluating gives approximately: - For \( t = 0.0001 \), value is 5.00050.- For \( t = -0.0001 \), value is 4.99950.

Step 7: Observing the Trend

The function values converge to approximately 5 as \( t \to 0 \) from both positive and negative directions.

Key Concepts

Exponential Function

The exponential function is a fundamental mathematical concept that is widely used to model growth and decay processes. In mathematical terms, the exponential function denoted by \( e^x \) is defined as a series expansion:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]This series continues indefinitely, but for small values of \( x \), the terms decrease rapidly, and the function can be approximated by just a few terms.

In the problem given, the function is \( e^{5t} \), which is an exponential function raised to the power of \( 5t \). This results in more pronounced changes as compared to \( e^t \) especially for larger \( t \) values. When dealing with limits, the behavior of the exponential function near zero is critical as it defines how sharply an exponential function either expands or contracts.

l'Hôpital's Rule

When solving limits of forms like \( \frac{0}{0} \), l'Hôpital's Rule becomes very useful. This rule provides a method to evaluate indeterminate forms by differentiation.

The rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \), then:

• We can differentiate the numerator and the denominator separately and find:

\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]Provided that the limit on the right exists. For our specific limit problem \( \lim_{t \to 0} \frac{e^{5t} - 1}{t} \), we have an indeterminate form. Here, l'Hôpital's Rule can be applied by finding the derivatives:

• Derivative of the numerator \( e^{5t} - 1 \), which is \( 5e^{5t} \).
• Derivative of the denominator \( t \), which is \( 1 \).

Applying l'Hôpital's, the limit simplifies to \( 5 \cdot e^{5 \times 0} = 5 \), confirming the trend found in our numerical approximations.

Numerical Approximation

Numerical approximation is a method used to calculate an approximate solution rather than an exact answer. It's especially handy when functions are complex or when limits are difficult to evaluate directly.

In the exercise given, we approximate the limit of the expression by evaluating the function at values of \( t \) that are increasingly close to zero, both from the positive and negative sides. By calculating:

• For \( t = 0.5 \), \( 0.1 \), \( 0.01 \), \( 0.001 \), and \( 0.0001 \), both positive and negative values are evaluated.
• We see convergence trends, and the function values get closer to 5 as \( t \) approaches zero.

This approach supports the hypothesis about the limit's value as well as offers insight into the behavior of the function near the limit point.

Convergence of Sequences

Convergence of sequences is a fundamental concept in calculus that involves understanding how the terms of a sequence behave as they go to infinity or approach a specific value. This concept is key when observing the behavior of functions and their limits.

In our evaluation of \( \lim_{t \to 0} \frac{e^{5t} - 1}{t} \), we observe a sequence of function values at different \( t \) values. The values of this sequence for both approaching from the positive and the negative sides are crucial. They inform us if the sequence converges to a single finite value as \( t \) comes close to zero.

• From our step-by-step evaluations, the function values appear to stabilize around 5.
• This stabilization indicates the sequence is converging towards the limit of 5.

Understanding this convergence can reinforce the hypothesis drawn from the numerical approximations and algebraic evaluations, and it underscores the fundamental properties of limits in calculus.