Problem 36 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Check the Form of the Limit

To use L'Hôpital's rule, the limit must initially be in an indeterminate form, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Evaluate the limit \( \lim _{y \rightarrow 0} \frac{\sqrt{a y+a^{2}}-a}{y} \). When \( y = 0 \), the expression becomes \( \frac{\sqrt{a \cdot 0 + a^2} - a}{0} = \frac{a - a}{0} = \frac{0}{0} \). This is an indeterminate form.

Step 2: Apply L'Hôpital's Rule

Since the limit is in the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that \( \lim _{x \rightarrow c} \frac{f(x)}{g(x)} = \lim _{x \rightarrow c} \frac{f'(x)}{g'(x)} \), provided the latter limit exists. Differentiate the numerator and the denominator separately with respect to \( y \).

Step 3: Differentiate the Numerator

The numerator is \( \sqrt{a y + a^2} - a \). Differentiate this with respect to \( y \). The derivative of \( \sqrt{a y + a^2} \) is \( \frac{1}{2\sqrt{a y + a^2}} \cdot a \) using the chain rule. The derivative of \( a \) is \( 0 \). Thus, the derivative of the numerator is \( \frac{a}{2\sqrt{a y + a^2}} \).

Step 4: Differentiate the Denominator

The denominator is simply \( y \). Differentiate this with respect to \( y \), which gives \( 1 \).

Step 5: Use L'Hôpital's Rule to Compute the Limit

Now apply L'Hôpital's Rule by substituting the derivatives into the limit: \( \lim _{y \rightarrow 0} \frac{\frac{a}{2\sqrt{a y + a^2}}}{1} = \lim _{y \rightarrow 0} \frac{a}{2\sqrt{a y + a^2}} \).

Step 6: Evaluate the Limit

Substitute \( y = 0 \) into the expression \( \frac{a}{2\sqrt{a y + a^2}} \) to find the limit: \( \frac{a}{2\sqrt{a \cdot 0 + a^2}} = \frac{a}{2a} = \frac{1}{2} \).

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form arises when performing certain limit evaluations. You're often left with expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), which at face value don't give you enough information to determine a limit. The concept of indeterminate forms captures this uncertainty. Think of it as a signpost indicating that deeper mathematical tools are needed to resolve the ambiguity. This happens often in situations involving fractions, where both the numerator and the denominator approach zero or infinity.

Recognizing indeterminate forms is the first crucial step before applying more advanced techniques, like L'Hôpital's Rule. This step assures that the initial results don't immediately define the limit, prompting further evaluation.

Limit Evaluation

Limit evaluation is a fundamental part of calculus, pivotal for understanding the behavior of functions as variables approach certain values. By evaluating limits, you estimate the value a function is trending toward even if the function doesn't actually reach that point. In our exercise, we needed to evaluate the limit of \( \lim_{y \rightarrow 0} \frac{\sqrt{a y+a^{2}}-a}{y} \). Initially, our expression simplified to the form \( \frac{0}{0} \), indicating indeterminacy.

Here, L'Hôpital's Rule becomes an incredibly useful tool. This rule allows us to take derivatives of the numerator and denominator separately, transforming an indeterminate form into a solvable one. It's especially handy when other algebraic approaches seem cumbersome. In our case, applying L'Hôpital's Rule clarified the expression greatly, leading us to a computable form and eventually to the answer.

Differentiation Process

Differentiation is the process by which we find the derivative of a function. It measures how a function's output changes with respect to changes in the input. In the context of L'Hôpital's Rule, you apply differentiation to both the numerator and denominator separately before re-evaluating the limit.

In our scenario, differentiating the numerator \( \sqrt{a y + a^2} - a \) involved using the chain rule. This rule is vital for differentiating composite functions efficiently. For the term \( \sqrt{a y + a^2} \), we considered its derivative as \( \frac{1}{2\sqrt{a y + a^2}} \times a \). Meanwhile, the derivative of the denominator \( y \) was straightforward: simply 1.

• The numerator's differentiation required careful application of the chain rule, multiplying by the derivative of the inner function.
• The denominator's derivative, being linear, simplified the process significantly.

Once differentiated, the expression becomes manageable, and re-evaluating the limit yields the correct solution, illustrating the power of these calculus tools.