Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Evaluate the Following Limit**
\[
\lim_{{x \to -6}} \frac{{x^2 - 36}}{{x + 6}}
\]
**Explanation:**
The expression involves finding the limit of a rational function as \( x \) approaches \(-6\). The numerator is a quadratic expression \( x^2 - 36 \), which can be factored as the difference of squares:
\[ x^2 - 36 = (x - 6)(x + 6) \]
By factoring the numerator, the expression becomes:
\[ \frac{{(x - 6)(x + 6)}}{{x + 6}} \]
You'll notice that \( (x + 6) \) appears in both the numerator and the denominator, allowing you to cancel it out, given that \( x \neq -6 \):
\[ = x - 6 \]
Thus, the limit simplifies to evaluating the expression \( x - 6 \) as \( x \to -6 \):
\[ \lim_{{x \to -6}} (x - 6) = -6 - 6 = -12 \]
Therefore, the limit is \(-12\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17a10252-ae51-4d74-b5c4-8b720ae84599%2Ff27d8b0c-1c81-45d8-b13f-3a6e3c5e65ed%2Fl47vh6tr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Evaluate the Following Limit**
\[
\lim_{{x \to -6}} \frac{{x^2 - 36}}{{x + 6}}
\]
**Explanation:**
The expression involves finding the limit of a rational function as \( x \) approaches \(-6\). The numerator is a quadratic expression \( x^2 - 36 \), which can be factored as the difference of squares:
\[ x^2 - 36 = (x - 6)(x + 6) \]
By factoring the numerator, the expression becomes:
\[ \frac{{(x - 6)(x + 6)}}{{x + 6}} \]
You'll notice that \( (x + 6) \) appears in both the numerator and the denominator, allowing you to cancel it out, given that \( x \neq -6 \):
\[ = x - 6 \]
Thus, the limit simplifies to evaluating the expression \( x - 6 \) as \( x \to -6 \):
\[ \lim_{{x \to -6}} (x - 6) = -6 - 6 = -12 \]
Therefore, the limit is \(-12\).
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