Problem 103 - Calculus and its applications

Question

Find the function values that are approximations for e. Round to five decimal places. $$ \begin{aligned} &\text { For } g(t)=t^{1 /(t-1)}, \text { we have } e=\lim _{t \rightarrow 1} g(t) . \text { Find } g(0.5)\\\ &g(0.9), g(0.99), g(0.999), \text { and } g(0.9998) \end{aligned} $$

Step-by-Step Solution

Answer

The approximate values of \( g(t) \) as \( t \) approaches 1 are 4, 2.65614, 2.70481, 2.71692, and 2.71814.

Step 1: Understanding the Limit Expression

We are given the function \( g(t) = t^{1/(t-1)} \) and need to evaluate it close to \( t = 1 \) because \( e = \lim_{t \rightarrow 1} g(t) \). We want to find the approximate value of \( e \) by calculating \( g(t) \) for values of \( t \) near 1.

Step 2: Calculating g(0.5)

Substitute \( t = 0.5 \) into the function: \[g(0.5) = (0.5)^{1/(0.5 - 1)} = (0.5)^{-2} = 4.\]

Step 3: Calculating g(0.9)

Substitute \( t = 0.9 \) into the function:\[g(0.9) = (0.9)^{1/(0.9 - 1)} = (0.9)^{-10} = 2.65614 \] (rounded to five decimal places).

Step 4: Calculating g(0.99)

Substitute \( t = 0.99 \) into the function: \[g(0.99) = (0.99)^{1/(0.99 - 1)} = (0.99)^{-100} = 2.70481 \] (rounded to five decimal places).

Step 5: Calculating g(0.999)

Substitute \( t = 0.999 \) into the function:\[g(0.999) = (0.999)^{1/(0.999 - 1)} = (0.999)^{-1000} = 2.71692 \] (rounded to five decimal places).

Step 6: Calculating g(0.9998)

Substitute \( t = 0.9998 \) into the function:\[g(0.9998) = (0.9998)^{1/(0.9998 - 1)} = (0.9998)^{-5000} = 2.71814 \] (rounded to five decimal places).

Step 7: Analyzing the Limit Values

By observing the calculated values of \( g(t) \) as \( t \) approaches 1, we see them getting closer to the natural logarithm base \( e \), approximately 2.71828.

Key Concepts

Exponential Approximations

Exponential approximations are useful when evaluating functions, especially when approaching limits. Such approximations involve understanding how an exponential function behaves near a particular value. In the exercise, the function is in the form of an exponential expression, where the base and the exponent play crucial roles. Here, we explore the function \( g(t) = t^{1/(t-1)} \). This expression becomes significant because as \( t \) approaches 1, the function approximates the natural number \( e \), a fundamental constant in mathematics. To approximate \( e \), we substitute values of \( t \) getting increasingly closer to 1, such as 0.5, 0.9, 0.99, etc. Each substitution provides a closer approximation of \( e \), utilizing the exponential nature of the given function. In practical scenarios, exponential approximations are beneficial:

• to simplify complex calculations,
• in numerical methods where exact computation is challenging, and
• in various fields such as finance, physics, and engineering.

So, embracing how exponential approximations work can empower learners to better tackle other complex problems.

Limits and Approaching Values

Limits are a cornerstone of calculus, providing a foundation for understanding how functions behave near a specific point. In the exercise, the limit \( \lim_{t \rightarrow 1} g(t) \) is key to approximating \( e \). This involves analyzing the behavior of \( g(t) = t^{1/(t-1)} \) as \( t \) gets closer and closer to 1. Limits help us predict the behavior of a function around a point without necessarily reaching that point. In our scenario, though \( t \) never actually equals 1, the function's values hint strongly at what value the function should have if \( t \) were 1. This is exactly how limits enable us to understand functions in calculus.The values \( g(0.5), g(0.9), g(0.99), g(0.999), \) and \( g(0.9998) \) give us a sequence that increasingly approaches \( e \). Understanding limits allows us to see these values converge to \( e \), illustrating the principle that as \( t \) "approaches" 1, the function's result closes in on \( e \). Understanding limits and approaching values is essential for:

• solving continuity problems,
• analyzing instantaneous rates of change, and
• finding derivatives and integrals in calculus.

Grasping this concept is vital for progressing in mathematical studies and applying calculus in practical situations.

Evaluating Functions Near a Point

Evaluating functions near a specific point involves substituting values close to this point and observing how the function behaves. This approach is particularly useful in calculus when exact values may be challenging to compute or when exploring limits and approximations.In our exercise, we examine the function \( g(t) = t^{1/(t-1)} \) near \( t = 1 \). By choosing values of \( t \) close to 1, like 0.5, 0.9, 0.99, etc., we observe how the function \( g(t) \) approaches the value of \( e \). When evaluating such functions near a point, sometimes:

• The function may exhibit behavior such as continuity or discontinuity at the point.
• The behavior at these points can differ significantly from the behavior elsewhere.

For instance, as \( t \) approaches 1, \( g(t) \) yields values converging towards \( e \), showcasing how even slight changes in \( t \) can influence the function's outcome. Such evaluations are critical because they:

• allow us to understand the immediate behavior of complex functions,
• assist in graphical interpretations of functions,
• and are foundational for the study of calculus related topics like derivatives and integrals.

Learning to evaluate functions near a point strengthens one's problem-solving capabilities in mathematics.