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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Find the limit, if it exist. (Use Squeeze Theorem)
![**Finding the Limit Using the Squeeze Theorem**
**Objective:**
Determine whether the limit exists, and if it does, find its value by using the Squeeze Theorem.
**Problem Statement:**
\[ \lim_{{(x,y) \to (0,0)}} \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \]
### Explanation:
To apply the Squeeze Theorem to find the limit, we'll start by analyzing the given function and simplifying it under constraints.
### Step-by-Step Solution:
1. **Understand the Function:**
The given function is:
\[ f(x,y) = \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \]
2. **Bound the Function:**
Recognize that \(\sin^2(y)\) is always between 0 and 1 for all real values of \(y\). Therefore:
\[ 0 \leq \sin^2(y) \leq 1 \]
3. **Multiplying Inequalities:**
Because \(\sin^2(y)\) is bounded between 0 and 1:
\[ 0 \leq x^2 \sin^2(y) \leq x^2 \]
4. **Formulate Bounding Fractions:**
Thus,
\[ 0 \leq \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \leq \frac{x^2}{x^2 + 5y^2} \]
5. **Analyze Bounding Values:**
\[ \frac{x^2}{x^2 + 5y^2} \]
Since both \(x\) and \(y\) are approaching 0, the numerator \(x^2\) also approaches 0. The denominator \(x^2 + 5y^2\) is positive and approaches 0 but is always greater than or equal to \(x^2\).
6. **Apply the Squeeze Theorem:**
\[ 0 \leq \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \leq \frac{x^2}{x^2 + 5y^2} \]
As \((x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa824ef54-0ee2-4591-9058-ff8a94975446%2F1dc880f7-a664-41a3-ad8b-1ba9ef71656b%2Fwgwm3bm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Limit Using the Squeeze Theorem**
**Objective:**
Determine whether the limit exists, and if it does, find its value by using the Squeeze Theorem.
**Problem Statement:**
\[ \lim_{{(x,y) \to (0,0)}} \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \]
### Explanation:
To apply the Squeeze Theorem to find the limit, we'll start by analyzing the given function and simplifying it under constraints.
### Step-by-Step Solution:
1. **Understand the Function:**
The given function is:
\[ f(x,y) = \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \]
2. **Bound the Function:**
Recognize that \(\sin^2(y)\) is always between 0 and 1 for all real values of \(y\). Therefore:
\[ 0 \leq \sin^2(y) \leq 1 \]
3. **Multiplying Inequalities:**
Because \(\sin^2(y)\) is bounded between 0 and 1:
\[ 0 \leq x^2 \sin^2(y) \leq x^2 \]
4. **Formulate Bounding Fractions:**
Thus,
\[ 0 \leq \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \leq \frac{x^2}{x^2 + 5y^2} \]
5. **Analyze Bounding Values:**
\[ \frac{x^2}{x^2 + 5y^2} \]
Since both \(x\) and \(y\) are approaching 0, the numerator \(x^2\) also approaches 0. The denominator \(x^2 + 5y^2\) is positive and approaches 0 but is always greater than or equal to \(x^2\).
6. **Apply the Squeeze Theorem:**
\[ 0 \leq \frac{x^2 \sin^2(y)}{x^2 + 5y^2} \leq \frac{x^2}{x^2 + 5y^2} \]
As \((x
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