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Magazine Chapter 2: Problem 31
Find the limits. $$\lim _{y \rightarrow 6^{+}} \frac{y+6}{y^{2}-36}$$
Short Answer
Expert verified
The limit is \(+\infty\).
Step by step solution
01
Analyze the Limit Expression
We are given the expression \( \lim _{y \rightarrow 6^{+}} \frac{y+6}{y^{2}-36} \). This indicates that we need to find the limit of the function \( f(y) = \frac{y+6}{y^{2}-36} \) as \( y \) approaches 6 from the right (positive side).
02
Simplify the Denominator
The denominator \( y^2 - 36 \) can be factored. Notice that \( 36 = 6^2 \), so \( y^2 - 36 \) is a difference of squares, which factors to \( (y - 6)(y + 6) \). Thus, our function becomes \( f(y) = \frac{y+6}{(y-6)(y+6)} \).
03
Cancel Common Factors
In the expression \( \frac{y+6}{(y-6)(y+6)} \), we see a common factor of \( y+6 \) in both the numerator and the denominator. We cancel \( y+6 \) from both, which simplifies the function to \( f(y) = \frac{1}{y-6} \) as long as \( y eq -6 \).
04
Evaluate the Limit as y Approaches 6 from the Right
Now that we have \( f(y) = \frac{1}{y-6} \), we evaluate the limit \( \lim_{y \rightarrow 6^+} f(y) \). As \( y \rightarrow 6^+ \), \( y - 6 \) approaches 0 from the positive side. Therefore, \( \frac{1}{y-6} \) approaches \( +\infty \).
05
Conclude the Solution
The limit \( \lim_{y \rightarrow 6^+} \frac{y+6}{y^2-36} \) equals \( +\infty \), indicating that the function grows unbounded positively as \( y \to 6^+ \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference of Squares
In calculus, the difference of squares is a useful algebraic identity that helps simplify complex expressions. The difference of squares takes the form \(a^2 - b^2\), and factors into \((a-b)(a+b)\). This is an essential concept used throughout mathematics to simplify polynomial expressions.
In the provided exercise, you have the expression \(y^2 - 36\). Recognizing this as a difference of squares is crucial. Here, \(a^2 = y^2\) and \(b^2 = 36\), which means \(b = 6\). Thus, the difference of squares formula tells us that \(y^2 - 36\) equals \((y - 6)(y + 6)\).
Understanding how to apply the difference of squares will enhance your algebraic manipulation skills and improve problem-solving efficiency.
In the provided exercise, you have the expression \(y^2 - 36\). Recognizing this as a difference of squares is crucial. Here, \(a^2 = y^2\) and \(b^2 = 36\), which means \(b = 6\). Thus, the difference of squares formula tells us that \(y^2 - 36\) equals \((y - 6)(y + 6)\).
- Helps simplify the denominator by factoring.
- Reveals common factors with the numerator, allowing further simplification.
- Makes evaluating limits more straightforward.
Understanding how to apply the difference of squares will enhance your algebraic manipulation skills and improve problem-solving efficiency.
Factoring Polynomials
Factoring is the process of breaking down a complex polynomial into simpler components called factors. This is particularly helpful when solving equations and evaluating limits by highlighting common terms.
In the given problem, after identifying \(y^2 - 36\) as a difference of squares, you factor it into \((y - 6)(y + 6)\). Factoring transforms the function to \(\frac{y+6}{(y-6)(y+6)}\).
You can then cancel out the \(y+6\) terms, assuming \(y eq -6\), simplifying the function to \(\frac{1}{y-6}\). This step is crucial for evaluating limits, as it reduces the complexity of the function drastically, making direct substitutions possible at some points.
Mastering polynomial factoring not only helps with limits but also with solving equations and sketching polynomial functions.
In the given problem, after identifying \(y^2 - 36\) as a difference of squares, you factor it into \((y - 6)(y + 6)\). Factoring transforms the function to \(\frac{y+6}{(y-6)(y+6)}\).
You can then cancel out the \(y+6\) terms, assuming \(y eq -6\), simplifying the function to \(\frac{1}{y-6}\). This step is crucial for evaluating limits, as it reduces the complexity of the function drastically, making direct substitutions possible at some points.
- Reduces polynomials to simpler forms.
- Identifies and removes common terms.
- Facilitates limit evaluation in calculus problems.
Mastering polynomial factoring not only helps with limits but also with solving equations and sketching polynomial functions.
One-Sided Limits
When finding limits in calculus, one-sided limits allow you to consider the behavior of a function as the input approaches a specific value from one side only. This is especially useful when dealing with points of discontinuity or infinite behavior.
In this limit problem, \(\lim_{y \rightarrow 6^+} \frac{1}{y-6}\), you're examining the function as \(y\) approaches 6 from values greater than 6. The notation \(6^+\) specifies approaching from the right side, which means that \(y\) is getting closer to 6 through values like 6.1, 6.01, 6.001, etc.
As \(y\) approaches 6 from the right, \(y - 6\) approaches 0 from the positive side, making \(\frac{1}{y-6}\) shoot towards \(+\infty\).
Grasping one-sided limits is key to mastering the concept of continuity and the behavior of functions around points of interest.
In this limit problem, \(\lim_{y \rightarrow 6^+} \frac{1}{y-6}\), you're examining the function as \(y\) approaches 6 from values greater than 6. The notation \(6^+\) specifies approaching from the right side, which means that \(y\) is getting closer to 6 through values like 6.1, 6.01, 6.001, etc.
As \(y\) approaches 6 from the right, \(y - 6\) approaches 0 from the positive side, making \(\frac{1}{y-6}\) shoot towards \(+\infty\).
- Helps understand directional limits at a point.
- Clarifies behavior at discontinuities.
- Essential for understanding asymptotic behavior.
Grasping one-sided limits is key to mastering the concept of continuity and the behavior of functions around points of interest.
