Problem 54 - Calculus for Biology and Medicine

Question

Compute the limits in Problems \(53-56 .\) $$ \lim _{h \rightarrow 0} \frac{e^{5 h}-1}{3 h} $$

Step-by-Step Solution

Answer

The limit is \( \frac{5}{3} \).

Step 1: Recognize the Limit Form

The given limit is \[\lim _{h \rightarrow 0} \frac{e^{5 h}-1}{3 h} \]This is a form that can be approached using the property of derivatives or limits involving exponential functions. Specifically, it resembles the derivative of the exponential function at zero.

Step 2: Apply Limit Properties

Recognize that \[\lim_{x \to 0} \frac{e^x - 1}{x} = 1\]Given this, if we substitute \(x = 5h\), we need to rewrite the limit:\[\lim_{h \to 0} \frac{e^{5h}-1}{3h} = \frac{1}{3} \cdot \lim_{h \to 0} \frac{e^{5h}-1}{5h} \cdot 5\]

Step 3: Simplify Using Known Limits

Based on the properties of limits, the following transformation can be used:\[\lim_{x \to 0} \frac{e^x - 1}{x} = 1\]So, the limit transforms to:\[\lim_{h \to 0} \frac{1}{3} \cdot 5 \cdot 1 = \frac{5}{3}\]

Step 4: Conclude the Result

Putting it all together, we find that:\[\lim _{h \rightarrow 0} \frac{e^{5 h}-1}{3 h} = \frac{5}{3}\]This is the simplified answer to the problem using the properties of exponential limits.

Key Concepts

Exponential Functions

Exponential functions are one of the fundamental building blocks in mathematics, especially in calculus. They take the form \( f(x) = a^x \) where \( a \) is a constant, and \( x \) is the variable. A very special case is when the base \( a \) is Euler's number \( e \), approximately equal to 2.71828. This results in the function \( e^x \), which has unique and useful properties.
Here are some interesting points about exponential functions:

• Growth: Exponential functions exhibit rapid growth compared to polynomial functions. As \( x \) increases, \( e^x \) grows faster than any power of \( x \).
• Decay: When negative exponents are involved, such as \( e^{-x} \), the function models exponential decay.
• Continuity and Differentiability: The function \( e^x \) is continuous and differentiable at every point in its domain, making it very useful in calculus.

Understanding exponential functions helps in grasping many concepts in calculus, particularly when dealing with limits and derivatives, as they often involve exponential expressions.

Derivative of Exponential Function

Let's dive into how derivatives apply to exponential functions, especially the base \( e \). The derivative of the exponential function \( e^x \) is quite remarkable because it produces itself. In other words, \( \frac{d}{dx}(e^x) = e^x \).
This unique property simplifies the process of finding derivatives involving exponential terms. Here's how it works with a constant multiplier, such as we often see in calculus problems:

• Derivative Rule for Exponential Functions: If you have \( f(x) = e^{kx} \), then the derivative is \( f'(x) = ke^{kx} \), following the chain rule.
• Common Application: It frequently applies to problems involving rates of change, compound interest, and growth/decay models.

Understanding the derivative of exponential functions is crucial. It forms the basis of solving various problems in calculus, like those involving natural exponential limits.

Limit Properties

Limits are foundational in calculus, providing insights into the behavior of functions as they approach specific points. The problem we are looking at involved using limits in combination with exponential functions. Here’s a breakdown of key limit properties that are helpful:

• Basic Limit: One of the fundamental limits to know is \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \). This simplifies calculations involving exponential functions near zero.
• Substitution and Multiplication: When variables are scaled, such as replacing \( x \) with \( 5h \), the resulting adjustments must be made in the expression \( \frac{e^x - 1}{x} \) by scaling appropriately, often involving multiplication or division by constants.
• Combining Limits: Limits can often be broken into manageable parts. If you recognize a piece of the function that matches a known limit, substitute directly. This is similar to our solution where the problem was transformed using \( \frac{1}{3} \cdot \lim_{h \to 0} \frac{e^{5h}-1}{5h} \cdot 5\).

Understanding how to manipulate and apply limit properties allows for effective solution strategies in calculus, especially problems involving forms like \( \frac{0}{0} \). It helps you simplify complex expressions neatly.