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Magazine Chapter 2: Problem 11
Find the value of the limit and when applicable indicate the limit theorems being used.\(\lim _{y \rightarrow-3} \sqrt{\frac{y^{2}-9}{2 y^{2}+7 y+3}}\)
Short Answer
Expert verified
The limit is \(\frac{\sqrt{30}}{5}\).
Step by step solution
01
Identify Direct Substitution
To solve the limit \(\lim_{y\rightarrow-3} \sqrt{\frac{y^{2}-9}{2 y^{2}+7 y+3}}\), first check if direct substitution works. Substitute \(-3\) into the expression: \(\frac{(-3)^{2}-9}{2(-3)^{2}+7(-3)+3} = \frac{9-9}{18-21+3} = \frac{0}{0}\). Since this results in an indeterminate form, further steps are needed.
02
Simplify the Expression
Simplify the expression inside the limit. Factor the numerator and denominator: \(y^{2}-9 = (y+3)(y-3)\) and \(2 y^{2}+7 y+3 = (2y+1)(y+3)\). Thus, the limit becomes: \(\frac{(y+3)(y-3)}{(2y+1)(y+3)}\).
03
Cancel Common Terms
Cancel the common \(y+3\) term from the numerator and denominator: \(\frac{(y-3)}{(2y+1)}\). The expression simplifies to: \(\frac{y-3}{2y+1}\).
04
Reevaluate the Limit
Substitute \(-3\) into the simplified expression: \(\frac{-3-3}{2(-3)+1} = \frac{-6}{-5} = \frac{6}{5}\).
05
Apply the Square Root
Apply the square root to the fraction: \(\frac{6}{5}\). Thus, \(\sqrt{\frac{6}{5}}\). This can be written as \(\frac{\sqrt{6}}{\sqrt{5}} = \frac{\sqrt{30}}{5}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Theorems
When calculating limits, we often use a variety of theorems to simplify and evaluate expressions. Here are some important ones:
- Sum/Difference Theorem: This states that the limit of a sum (or difference) is the sum (or difference) of the limits. For example, \(\text{ if } \lim_{y \to a} f(y) = L \text{ and } \lim_{y \to a} g(y) = M, \text{ then } \lim_{y \to a} (f(y) + g(y)) = L + M.\)
- Product Theorem: The limit of a product is the product of the limits. For instance, \(\text{ if } \lim_{y \to a} f(y) = L \text{ and } \lim_{y \to a} g(y) = M, \text{ then } \lim_{y \to a} (f(y) \times g(y)) = L \times M.\)
- Quotient Theorem: The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero. Thus, \(\text{ if } \lim_{y \to a} f(y) = L \text{ and } \lim_{y \to a} g(y) = M eq 0, \text{ then } \lim_{y \to a} \frac{ f(y) }{ g(y) } = \frac{ L }{ M }\).
Indeterminate Forms
Indeterminate forms like \( \frac{0}{0}\) or \(\frac{\text{鈭瀩}{\text{鈭瀩}\) arise during limit evaluation and indicate that more work is needed to simplify the expression. Let's delve deeper:
- What is \(\frac{0}{0}\) Form? This occurs when both the numerator and denominator of a fraction approach zero as we take the limit. It's indeterminate because zero divided by zero doesn't yield a finite or clear value.
- Resolving \(\frac{0}{0}\): To resolve such forms, we use techniques like algebraic simplification, L'Hopital's Rule, or factorization. For example, in our exercise, substituting \(-3\) into the given limit resulted in \(\frac{0}{0}\), prompting further steps to simplify the expression.
Simplifying Rational Expressions
A rational expression is a fraction where both the numerator and the denominator are polynomials. Simplifying these expressions is crucial for evaluating limits:
- Factorization: Identify and factor common terms in the numerator and denominator. For example, \( \frac{y^2 - 9}{2y^2 + 7y + 3}\) was factored into \( \frac{(y+3)(y-3)}{(2y+1)(y+3)}\).
- Canceling Common Terms: Once factored, cancel out identical terms in the numerator and denominator. In our example, the \( (y+3)\) terms were canceled, simplifying the expression to \(\frac{y-3}{2y+1}\).
Factorization
Factorization is the process of breaking down a complex expression into simpler, multipliable components. It's foundational for simplifying and solving limits:
- Difference of Squares: Recognize patterns like \(a^2 - b^2 = (a + b)(a - b)\). In the exercise, \(y^2 - 9\) was rewritten as \((y+3)(y-3)\).
- Quadratic Trinomials: Factor expressions of the form \(ax^2 + bx + c\) using methods like the quadratic formula or simple trial and error. Here, \(2y^2 + 7y + 3\) was factorable into \((2y+1)(y+3)\).
