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
use the speical limits please if possible and if the limit doesn't exist please explain why.
![### Understanding Limits with Trigonometric Functions
Let's evaluate the following limit:
\[ \lim_{{x \to -3}} \frac{2 \sin(x + 3)}{x^2 - x - 12} \]
This expression involves finding the limit of the function as \( x \) approaches \(-3\). Here, we see a trigonometric function in the numerator and a quadratic polynomial in the denominator.
First, we will look at each part of the expression separately:
#### Numerator:
\[ 2 \sin(x + 3) \]
This is simply twice the sine of \((x + 3)\).
#### Denominator:
\[ x^2 - x - 12 \]
This quadratic expression can be factored to:
\[ (x - 4)(x + 3) \]
Thus, the original limit can be rewritten as:
\[ \lim_{{x \to -3}} \frac{2 \sin(x + 3)}{(x - 4)(x + 3)} \]
To evaluate the limit, notice that the factor \((x + 3)\) in the numerator and denominator will cancel out:
\[ = \lim_{{x \to -3}} \frac{2 \sin(x + 3) / (x + 3)}{x - 4} \]
As \( x \) approaches \(-3\), the sine function can be approximated by its argument (as \(\sin a \approx a\) for small \( a \)):
\[ \approx \lim_{{x \to -3}} \frac{2(x + 3)/(x + 3)}{x - 4} \]
\[ = \lim_{{x \to -3}} \frac{2}{x - 4} \]
Now substitute \( x = -3 \):
\[ = \frac{2}{-3 - 4} \]
\[ = \frac{2}{-7} \]
\[ = -\frac{2}{7} \]
Therefore, the limit is:
\[ \lim_{{x \to -3}} \frac{2 \sin(x + 3)}{x^2 - x - 12} = -\frac{2}{7} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5ac30416-a057-4cd1-8bfb-7fbbdadcb376%2Fc7e6b360-662e-425a-b5b6-95be830adc9f%2F1qcruoo_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Limits with Trigonometric Functions
Let's evaluate the following limit:
\[ \lim_{{x \to -3}} \frac{2 \sin(x + 3)}{x^2 - x - 12} \]
This expression involves finding the limit of the function as \( x \) approaches \(-3\). Here, we see a trigonometric function in the numerator and a quadratic polynomial in the denominator.
First, we will look at each part of the expression separately:
#### Numerator:
\[ 2 \sin(x + 3) \]
This is simply twice the sine of \((x + 3)\).
#### Denominator:
\[ x^2 - x - 12 \]
This quadratic expression can be factored to:
\[ (x - 4)(x + 3) \]
Thus, the original limit can be rewritten as:
\[ \lim_{{x \to -3}} \frac{2 \sin(x + 3)}{(x - 4)(x + 3)} \]
To evaluate the limit, notice that the factor \((x + 3)\) in the numerator and denominator will cancel out:
\[ = \lim_{{x \to -3}} \frac{2 \sin(x + 3) / (x + 3)}{x - 4} \]
As \( x \) approaches \(-3\), the sine function can be approximated by its argument (as \(\sin a \approx a\) for small \( a \)):
\[ \approx \lim_{{x \to -3}} \frac{2(x + 3)/(x + 3)}{x - 4} \]
\[ = \lim_{{x \to -3}} \frac{2}{x - 4} \]
Now substitute \( x = -3 \):
\[ = \frac{2}{-3 - 4} \]
\[ = \frac{2}{-7} \]
\[ = -\frac{2}{7} \]
Therefore, the limit is:
\[ \lim_{{x \to -3}} \frac{2 \sin(x + 3)}{x^2 - x - 12} = -\frac{2}{7} \]
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