Evaluate the following limit. CSC X-1 Ilm Xx/2 cot x Select the correct cholce below and, if necessary, fill in the answer box to complete your choice CSC X-1 lim cot2 x because each function is continuous for all x. O A. Xx/2 (Simplify your answer.)
Evaluate the following limit. CSC X-1 Ilm Xx/2 cot x Select the correct cholce below and, if necessary, fill in the answer box to complete your choice CSC X-1 lim cot2 x because each function is continuous for all x. O A. Xx/2 (Simplify your answer.)
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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Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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PLEASE answer both. This is my last question of the month :(((
![### Evaluation of Limits in Trigonometric Functions
#### Problem Statement:
Evaluate the following limit:
\[
\lim_{{x \to \pi/2}} \frac{{\csc x - 1}}{{\cot^2 x}}
\]
#### Solution Choices:
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
**A.**
\[
\lim_{{x \to \pi/2}} \frac{{\csc x - 1}}{{\cot^2 x}} = \boxed{} \quad \text{because each function is continuous for all } x.
\]
*(Simplify your answer.)*
**B.**
The limit does not exist and is neither \(\infty\) nor \(-\infty\).
---
In solving the problem, you may need to recall the definitions and properties of the trigonometric functions involved:
- \(\csc x = \frac{1}{\sin x}\)
- \(\cot x = \frac{\cos x}{\sin x}\)
Considering the behavior of these trigonometric functions as \( x \) approaches \( \pi/2 \) could help in simplifying the limit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d6b9ce8-f79c-47f5-aca3-f390e26bdcec%2F9f1036be-be1f-4425-9df4-cd0ad069e2b7%2Fcsvlrla_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Evaluation of Limits in Trigonometric Functions
#### Problem Statement:
Evaluate the following limit:
\[
\lim_{{x \to \pi/2}} \frac{{\csc x - 1}}{{\cot^2 x}}
\]
#### Solution Choices:
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
**A.**
\[
\lim_{{x \to \pi/2}} \frac{{\csc x - 1}}{{\cot^2 x}} = \boxed{} \quad \text{because each function is continuous for all } x.
\]
*(Simplify your answer.)*
**B.**
The limit does not exist and is neither \(\infty\) nor \(-\infty\).
---
In solving the problem, you may need to recall the definitions and properties of the trigonometric functions involved:
- \(\csc x = \frac{1}{\sin x}\)
- \(\cot x = \frac{\cos x}{\sin x}\)
Considering the behavior of these trigonometric functions as \( x \) approaches \( \pi/2 \) could help in simplifying the limit.
![### Continuity of Rational Functions
#### Problem Statement
Determine the intervals on which the following function is continuous:
\[ f(x) = \frac{x + 3}{x^2 - 9} \]
#### Solution
The function is continuous on the interval(s) \[\_\_\_\_\_\_\_\_\_\_\_\_\] (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.)
---
In this problem, we are asked to determine the intervals on which the given rational function is continuous. For a rational function, continuity is ensured everywhere except at the values that cause the denominator to be zero, as these values make the function undefined.
To find where the function \( f(x) = \frac{x + 3}{x^2 - 9} \) is undefined, we need to solve for when the denominator is zero:
\[ x^2 - 9 = 0 \]
Solve for \( x \):
\[ x^2 = 9 \]
\[ x = \pm 3 \]
Thus, \( x = 3 \) and \( x = -3 \) are the points where the function is discontinuous. As a result, the function is continuous on the intervals that do not include these points:
In interval notation, the function is continuous on:
\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \]
So, the intervals should be entered in the box as:
\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d6b9ce8-f79c-47f5-aca3-f390e26bdcec%2F9f1036be-be1f-4425-9df4-cd0ad069e2b7%2Fkruvtmf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Continuity of Rational Functions
#### Problem Statement
Determine the intervals on which the following function is continuous:
\[ f(x) = \frac{x + 3}{x^2 - 9} \]
#### Solution
The function is continuous on the interval(s) \[\_\_\_\_\_\_\_\_\_\_\_\_\] (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.)
---
In this problem, we are asked to determine the intervals on which the given rational function is continuous. For a rational function, continuity is ensured everywhere except at the values that cause the denominator to be zero, as these values make the function undefined.
To find where the function \( f(x) = \frac{x + 3}{x^2 - 9} \) is undefined, we need to solve for when the denominator is zero:
\[ x^2 - 9 = 0 \]
Solve for \( x \):
\[ x^2 = 9 \]
\[ x = \pm 3 \]
Thus, \( x = 3 \) and \( x = -3 \) are the points where the function is discontinuous. As a result, the function is continuous on the intervals that do not include these points:
In interval notation, the function is continuous on:
\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \]
So, the intervals should be entered in the box as:
\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \]
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