Problem 39 - Calculus

Question

Investigate \(\lim _{h \rightarrow 0}(1+h)^{1 / h}\) numerically.

Step-by-Step Solution

Answer

The limit is approximately 2.718, which is \(e\).

Step 1: Understand the Limit

We want to find the numerical value of \( \lim _{h \rightarrow 0}(1+h)^{1 / h} \), which means we evaluate the expression \((1 + h)^{1/h}\) as \(h\) approaches 0.

Step 2: Choose Small Values of h

Start by choosing small positive and negative values of \(h\) such as 0.1, 0.01, -0.1, and -0.01 to observe the limit numerically.

Step 3: Calculate \((1+h)^{1/h}\) for Each h

Compute \((1+h)^{1/h}\) for each selected value of \(h\):- For \(h = 0.1\), calculate \((1.1)^{10}\).- For \(h = 0.01\), calculate \((1.01)^{100}\).- For \(h = -0.1\), calculate \((0.9)^{-10}\).- For \(h = -0.01\), calculate \((0.99)^{-100}\).

Key Concepts

Numerical Evaluation of Limits

Numerical evaluation of limits involves approximating the value of a limit by testing small values close to the point of interest. In this exercise, we are investigating how the expression \[ (1+h)^{1/h} \]behaves as \( h \) approaches zero. By choosing small positive and negative values for \( h \), such as 0.1, 0.01, -0.1, and -0.01, we can see how these changes affect the outcome. Small positive values for \( h \) allow us to observe the limit from the right side, while small negative values help us understand the behavior from the left.

• For \( h = 0.1 \), \( (1.1)^{10} \) is approximately 2.5937.
• For \( h = 0.01 \), \( (1.01)^{100} \) is approximately 2.7048.
• For \( h = -0.1 \), \( (0.9)^{-10} \) is approximately 2.5937.
• For \( h = -0.01 \), \( (0.99)^{-100} \) is approximately 2.7048.

The numerical evaluation helps us to observe the trend, which approaches a specific number as \( h \) gets closer to zero.

Exponential Function

The expression \[ (1+h)^{1/h} \]represents an exponential function, which is a type of function where a constant base is raised to a variable exponent. Exponential functions exhibit rapid growth or decay, depending on the base and the sign of the exponent. In this context, the base is slightly greater or lesser than 1, depending on the sign and magnitude of \( h \).
When \( h \) is positive, the expression represents an increasing exponential function, as seen in the cases

• \( h = 0.1 \)
• \( h = 0.01 \)

When \( h \) is almost zero but negative, the expression translates into a situation resembling exponential decay, since \( (1+h) \) becomes less than one in

• \( h = -0.1 \)
• \( h = -0.01 \)

This characteristic of exponential growth and decay is key to understanding why the evaluated numbers are so close in value despite different signs of \( h \).

Approaching Zero

Approaching zero is a fundamental concept in calculus, particularly when dealing with limits. In this exercise, "approaching zero" refers to making the values of \( h \) infinitesimally small, allowing us to explore what happens to our function\[ (1+h)^{1/h} \]as \( h \) nears zero. The idea is to find what value this expression gravitates towards as \( h \) gets smaller and smaller. As both small positive and negative \( h \) values were used:

• Positive \( h \) tests help to approach zero from the right.
• Negative \( h \) tests approach from the left.

By conducting this exploration, we can numerically see how values close to 0 generate outputs nearing an expected limit value.
This exercise demonstrates that, as \( h \) approaches zero, the expression converges to a value. Through this numerical observation, we can conclude that the limit \[ \lim _{h \rightarrow 0}(1+h)^{1 / h} \]is approximately \( e \), the base of natural logarithms, which emphasizes the unique behavior of exponential functions when approaching specific points.