L'Hôpital's Rule: 3 Practical Examples

Understanding L’Hôpital’s Rule

L’Hôpital’s Rule is a powerful tool in calculus used to evaluate limits of indeterminate forms such as 0/0 and ∞/∞. This rule states that if the limit of a function results in one of these forms, you can take the derivative of the numerator and the derivative of the denominator separately, then re-evaluate the limit. This method is particularly useful for simplifying complex expressions and is widely applicable in various mathematical problems.

Example 1: Evaluating a 0/0 Form

Context

Consider a scenario where we need to find the limit of a function as it approaches a certain point. This is common in physics and engineering when analyzing rates of change.

Limit:
\[ ext{Find } \lim_{x \to 0} \frac{\sin(x)}{x} \]\
As x approaches 0, both the numerator and denominator approach 0, creating a 0/0 indeterminate form.

Applying L’Hôpital’s Rule, we differentiate the numerator and denominator:

• Derivative of \( \sin(x) \) is \( \cos(x) \)
• Derivative of \( x \) is \( 1 \)

Now, we re-evaluate the limit:
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1 \]

Notes

This limit is essential in calculus and has applications in physics, particularly in wave motion and oscillations. A variation can be found in limits involving trigonometric functions approaching different angles.

Example 2: Evaluating an ∞/∞ Form

Context

In economics, we often deal with growth rates and limits that approach infinity. For instance, we might evaluate the behavior of a function describing profit as production increases indefinitely.

Limit:
\[ ext{Find } \lim_{x \to \infty} \frac{3x^3 + 2x^2}{5x^3 - 4x} \]\
As x approaches infinity, both the numerator and denominator tend toward infinity, resulting in an ∞/∞ form.

Applying L’Hôpital’s Rule involves differentiating:

• Derivative of \( 3x^3 + 2x^2 \) is \( 9x^2 + 4x \)
• Derivative of \( 5x^3 - 4x \) is \( 15x^2 - 4 \)

Re-evaluating the limit:
\[ \lim_{x \to \infty} \frac{3x^3 + 2x^2}{5x^3 - 4x} = \lim_{x \to \infty} \frac{9x^2 + 4x}{15x^2 - 4} = \frac{9}{15} = \frac{3}{5} \]

Notes

This example illustrates how L’Hôpital’s Rule can simplify limits involving polynomial functions. It also shows that higher-degree terms dominate in behavior as x approaches infinity.

Example 3: Evaluating a Complex Limit

Context

In scientific research, particularly in chemistry, one may encounter limits that involve exponential functions and logarithms. Understanding these limits can help in analyzing reaction rates.

Limit:
\[ ext{Find } \lim_{x \to 0} \frac{e^x - 1}{x} \]\
As x approaches 0, this limit results in a 0/0 form.

Using L’Hôpital’s Rule, we differentiate:

• Derivative of \( e^x - 1 \) is \( e^x \)
• Derivative of \( x \) is \( 1 \)

Re-evaluating the limit gives:
\[ \lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1 \]

Notes

This limit is vital in calculus and has implications in various fields, including economics and biology, where exponential growth is analyzed. Variations can include limits involving different bases of exponentials or applying L’Hôpital’s Rule multiple times for more complex functions.