Problem 30 - Thomas Calculus

Question

Limits of quotients Find the limits in Exercises \(23-42\) $$\lim _{y \rightarrow 0} \frac{5 y^{3}+8 y^{2}}{3 y^{4}-16 y^{2}}$$

Step-by-Step Solution

Answer

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

Step 1: Substitute Directly

Start by directly substituting the value of the limit, which is as \( y \) approaches 0, into the function: \[ \frac{5(0)^{3} + 8(0)^{2}}{3(0)^{4} - 16(0)^{2}} \] This simplifies to \( \frac{0}{0} \), which is an indeterminate form. Thus, we need to do more work to find the limit.

Step 2: Factor the Denominator

To resolve the indeterminate form, factor the denominator: \( 3y^{4} - 16y^{2} = y^{2}(3y^{2} - 16) \) This gives us a clearer view of potential cancellations with the numerator if it can also be factored or simplified.

Step 3: Simplify the Expression

Factor the numerator: \[ 5y^3 + 8y^2 = y^2(5y + 8) \]Thus, the expression becomes: \[ \frac{y^2(5y + 8)}{y^2(3y^2 - 16)} \] We can now cancel \( y^2 \) from both the numerator and the denominator, provided \( y eq 0 \).

Step 4: Evaluate the Limit of the Simplified Expression

After canceling \( y^2 \), the expression simplifies to: \[ \frac{5y + 8}{3y^2 - 16} \] Now, substitute \( y = 0 \) into this simplified expression to find the limit: \[ \lim_{y \rightarrow 0} \frac{5(0) + 8}{3(0)^2 - 16} = \frac{8}{-16} = -\frac{1}{2} \]

Step 5: Conclude the Solution

Now that we've evaluated the simplified expression as \( y \rightarrow 0 \), we conclude the limit: \[ \lim_{y \rightarrow 0} \frac{5y^3 + 8y^2}{3y^4 - 16y^2} = -\frac{1}{2} \]

Key Concepts

Indeterminate Forms

An indeterminate form is a mathematical expression that does not initially have a well-defined value. In calculus, these forms occur frequently when evaluating limits and can initially appear confusing. A common example is \( \frac{0}{0} \), which arises when substituting a value into a function leads to both the numerator and the denominator equating to zero.
To resolve an indeterminate form, we often need to re-evaluate or manipulate the function to find a clear result. Strategies like factoring, rationalizing, or transforming expressions using L'Hopital's Rule are typically employed. In the given exercise, after substituting \( y = 0 \) in the function, we ended up with \( \frac{0}{0} \). This is why further steps like factoring were necessary to determine the limit.

Factoring

Factoring is a technique used to simplify polynomials, which can make expressions easier to analyze and limits easier to calculate. The goal of factoring is to break down a complex expression into simpler components that can reveal possible simplifications.
In the provided solution, we factored both the numerator \( 5y^3 + 8y^2 \) and the denominator \( 3y^4 - 16y^2 \) as follows:

• The numerator was factored to \( y^2(5y + 8) \)
• The denominator was factored to \( y^2(3y^2 - 16) \)

By factoring, we were able to identify \( y^2 \) as a common term in both the numerator and denominator, which was then canceled out (assuming \( y eq 0 \)) to help simplify the expression for better evaluation of the limit.

Simplifying Expressions

Simplifying expressions is crucial for making complex algebraic expressions more manageable and can be particularly helpful when calculating limits. After factoring, the main objective is to cancel out terms that appear both in the numerator and the denominator, provided that canceling these terms does not inflate any undefined behaviors, such as division by zero.
In this exercise, after factoring, the expression: \[ \frac{y^2(5y + 8)}{y^2(3y^2 - 16)}\]was simplified by canceling out \( y^2 \), thus reducing the expression to:\[ \frac{5y + 8}{3y^2 - 16}\]Once simplified, substituting \( y = 0 \) into the equation results in a straightforward calculation of the limit \( -\frac{1}{2} \). This process of simplifying makes the evaluation of limits more straightforward by removing unnecessary complexity from the expression.