Problem 34 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{y \rightarrow 0} \frac{\sqrt{a y+a^{2}}-a}{y}, \quad a>0$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Evaluate the Indeterminate Form

First, we need to determine if the function \( \frac{\sqrt{a y + a^2} - a}{y} \) results in an indeterminate form when \( y \to 0 \). Substitute \( y = 0 \) into the numerator and denominator. The numerator becomes \( \sqrt{a \cdot 0 + a^2} - a = a - a = 0 \), and the denominator is \( y = 0 \). Thus, we have a \( \frac{0}{0} \) indeterminate form, allowing us to use l'Hôpital's Rule.

Step 2: Differentiate the Numerator and Denominator

Apply l'Hôpital's Rule, which requires differentiating the numerator and the denominator with respect to \( y \).The derivative of the numerator \( \sqrt{a y + a^2} - a \) is \( \frac{d}{dy}(\sqrt{a y + a^2}) = \frac{a}{2\sqrt{a y + a^2}} \).The derivative of the denominator \( y \) is 1.

Step 3: Apply l'Hôpital's Rule

After differentiating, we substitute back into the limit:\[ \lim_{y \to 0} \frac{a}{2\sqrt{a y + a^2}} = \frac{a}{2\sqrt{a \cdot 0 + a^2}} = \frac{a}{2a} = \frac{1}{2}. \]This gives us the limit of the original expression as \( y \to 0 \).

Key Concepts

Indeterminate Forms

When you are working with limits, you might encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms. They occur when both the numerator and denominator approach certain values that cause ambiguity in evaluation.
For instance, in the given exercise, substituting \( y = 0 \) results in a form \( \frac{0}{0} \). This signifies an uncertainty in determining the limit as both the top and bottom parts of the fraction equate to zero, making direct evaluation impossible.

• Indeterminate forms signal that a limit cannot be found by direct substitution alone.
• This is where techniques like l'Hôpital's Rule come into play.
• These forms include \( 0^0 \), \( \infty - \infty \), and many others, each requiring special handling to resolve.

Recognizing when you have an indeterminate form is crucial for applying the correct mathematical strategy to find a limit.

Limits

Limits are foundational in calculus, representing the value that a function approaches as the input approaches a certain point. In our context, we are trying to find \( \lim _{y \rightarrow 0} \frac{\sqrt{a y+a^{2}}-a}{y} \).
Limits help us understand the behavior of functions at points where they may not be explicitly defined by examining how outputs behave in infinitely close proximity.

• When calculating limits, substituting values directly into the function can be a first step.
• If it results in an indeterminate form, techniques like l'Hôpital's Rule are often used.
• Conceptually, it is about understanding what value a function is approaching, rather than what it exactly reaches.

The elegance of limits is in their ability to describe continuity and predict trends within the function, even in points of discontinuity or indeterminacy.

Differentiation

Differentiation is a process in calculus that involves finding the derivative of a function. This derivative represents the rate at which the function’s value changes as its input changes.
In l'Hôpital's Rule, differentiation is used to simplify \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) indeterminate forms by differentiating the numerator and the denominator separately.
For example, in the exercise, we found the derivative of \( \sqrt{a y + a^2} - a \) to be \( \frac{a}{2\sqrt{a y + a^2}} \).

• Differentiation helps in finding the slope or steepness of the tangent at any point on a curve.
• It's a crucial operation for applying l'Hôpital's Rule.
• Allows us to understand how functions grow or shrink.

This powerful mathematical tool transforms complex questions about functions into manageable calculations, revealing insights into how variables influence each other across entire domains.