Chapter 1: Problem 23
Find the limits. $$ \lim _{y \rightarrow 6} \frac{y+6}{y^{2}-36} $$
Short Answer
Expert verified
The limit does not exist (vertical asymptote at \( y = 6 \)).
Step by step solution
01
Identify the Expression
We need to find the limit: \( \lim_{y \rightarrow 6} \frac{y+6}{y^{2}-36} \). This expression involves a rational function, which is a fraction where both the numerator and denominator are polynomials.
02
Simplify the Denominator
The denominator \( y^2 - 36 \) is a difference of squares. We can factor it using the identity \( a^2 - b^2 = (a+b)(a-b) \). Thus:\[y^2 - 36 = (y + 6)(y - 6)\]
03
Substitute and Simplify
Now, substitute the factored form of the denominator into the limit expression:\[\lim_{y \rightarrow 6} \frac{y+6}{(y+6)(y-6)}\]We see that \( y + 6 \) appears in both the numerator and the denominator. Cancel out the common factor \( y + 6 \):\[\lim_{y \rightarrow 6} \frac{1}{y-6}\]
04
Evaluate the Limit
Now, evaluate the limit \( \lim_{y \rightarrow 6} \frac{1}{y-6} \). When \( y \) approaches 6, the expression \( y-6 \) in the denominator approaches 0.Since the denominator approaches zero and the numerator is a constant 1, the fraction goes towards infinity or negative infinity, indicating a vertical asymptote at \( y = 6 \).Hence, the limit does not exist.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Functions
A rational function is a type of function expressed as a fraction where both the numerator and the denominator are polynomials. In the world of mathematics, these functions are incredibly important for describing various real-world relationships. The given exercise is a perfect example of a rational function: \( \frac{y+6}{y^{2}-36} \). Here, "\( y+6 \)" serves as the numerator, and "\( y^{2}-36 \)" is the denominator. When dealing with rational functions, particularly in calculus, our primary concern is often about their limits, especially when a variable approaches a certain value. Rational functions can have a variety of behaviors based on the polynomials involved. One essential aspect is determining how the function behaves as the variable approaches a certain point, which can help identify discontinuities or asymptotic behaviors.
Factoring Polynomials
Factoring polynomials plays a critical role in simplifying complex rational functions. A polynomial is an expression involving variables and coefficients, utilizing operations such as addition and multiplication. In the present exercise, the polynomial involved in the denominator is \( y^{2} - 36 \).To factor this polynomial, identify it as a difference of squares. The difference of squares is a special case that can be factored using the identity: \( a^2 - b^2 = (a+b)(a-b) \). By recognizing \( y^{2} - 36 \) as \( (y+6)(y-6) \), we can simplify our given expression significantly. By factoring, you can cancel common factors from the numerator and the denominator, turning what seems like a complex limit into something much simpler. This method is particularly useful in determining limits and identifying removable discontinuities in rational functions.
Vertical Asymptotes
Vertical asymptotes are lines to which a graph of a function gets infinitely close as it approaches a certain value, but never touches or crosses. When evaluating limits in calculus, identifying vertical asymptotes is crucial, as they can indicate that a limit does not exist.In the exercise, after canceling the \( y+6 \) terms from the numerator and the factorized denominator, we reach the expression \( \frac{1}{y-6} \). As \( y \) approaches 6, the denominator approaches zero.Since the numerator is constantly 1, while the denominator gets very small, the fraction itself approaches infinity or negative infinity. These signs point toward a vertical asymptote at \( y = 6 \). The presence of a vertical asymptote signals that as we approach the given \( y \) value, the limit of the function does not exist, leading to either an infinite positive or negative value.