Problem 7
Question
\(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}\)
Step-by-Step Solution
Verified Answer
The limit is 1.
1Step 1: Identify the target limit
We are tasked with finding the limit of the function \( f(x, y) = \frac{\sin(x^{2}+y^{2})}{x^{2}+y^{2}} \) as \( (x, y) \to (0,0) \).
2Step 2: Convert to polar coordinates
In polar coordinates, \( x^{2} + y^{2} = r^{2} \), where \( r \) is the radius from the origin. So, the function becomes \( \frac{\sin(r^{2})}{r^{2}} \).
3Step 3: Apply known limit results
Recall the standard limit \( \lim_{r \to 0} \frac{\sin(r)}{r} = 1 \). This can be applied to our function by setting \( u = r^{2} \), which gives \( \lim_{r \to 0} \frac{\sin(r^{2})}{r^{2}} = \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \).
4Step 4: Conclude the limit value
As \( r \rightarrow 0 \), \( \frac{\sin(r^{2})}{r^{2}} \rightarrow 1 \). Therefore, the original limit \( \lim _{(x, y) \rightarrow(0,0)} \frac{\sin(x^{2}+y^{2})}{x^{2}+y^{2}} = 1 \).
Key Concepts
Polar CoordinatesLimits in Two VariablesTrigonometric Limits
Polar Coordinates
Polar coordinates are an alternative way of representing points in a plane, especially useful for systems involving circular motion or symmetry around a point. Instead of using Cartesian coordinates
In our multivariable calculus limit problem, converting \( x^2 + y^2 \) into \( r^2 \) simplifies the problem into a form that uses the symmetry of the circle. This makes it easier to work with functions and limits when \( r \to 0 \), focusing on the distance to the origin without worrying about the particular path.
- A point is defined by a distance from a fixed point (the origin), known as the radius (\( r \)).
- An angle from a fixed direction, often the positive x-axis, called the polar angle (\( \theta \)).
In our multivariable calculus limit problem, converting \( x^2 + y^2 \) into \( r^2 \) simplifies the problem into a form that uses the symmetry of the circle. This makes it easier to work with functions and limits when \( r \to 0 \), focusing on the distance to the origin without worrying about the particular path.
Limits in Two Variables
Limits in functions of two variables \( f(x, y) \) explore the behavior of \( f \) as \( (x, y) \) approaches some point, like \( (0, 0) \). These limits can be more complex than their single-variable counterparts due to multiple paths to approach the limit point.
Approaching \( (0,0) \) can be done along various paths, such as the x-axis, y-axis, or diagonals, leading to potentially different function values unless handled properly. Therefore, confirming a limit through different paths or using polar coordinates to reduce the multiple-variable problem to a single-variable form often helps in verifying consistency.
Approaching \( (0,0) \) can be done along various paths, such as the x-axis, y-axis, or diagonals, leading to potentially different function values unless handled properly. Therefore, confirming a limit through different paths or using polar coordinates to reduce the multiple-variable problem to a single-variable form often helps in verifying consistency.
- The polar conversion changes the variables into one parameter (\( r \)), making it easier to see that at \( r = 0 \), the resultant value is consistent no matter the direction of approach.
Trigonometric Limits
The exploration of trigonometric limits is a fundamental topic in calculus, deeply tied to understanding the behavior of functions involving trigonometric expressions. A classic result often used in calculus is the limit\( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \). This limit showcases how the sine function behaves around its origin and is pivotal for addressing many calculus problems involving sine.
In our specific limit problem, the formula \( \frac{\sin(u)}{u} \) serves as a critical tool for evaluating \( \frac{\sin(r^{2})}{r^{2}} \)as \( r \to 0 \). By setting \( u = r^2 \), we directly apply this standard limit to solve our multivariable problem. This reduces the complexity of evaluating the original expression, essentially shifting a two-variable problem to a familiar single-variable trigonometric limit.
In our specific limit problem, the formula \( \frac{\sin(u)}{u} \) serves as a critical tool for evaluating \( \frac{\sin(r^{2})}{r^{2}} \)as \( r \to 0 \). By setting \( u = r^2 \), we directly apply this standard limit to solve our multivariable problem. This reduces the complexity of evaluating the original expression, essentially shifting a two-variable problem to a familiar single-variable trigonometric limit.
- This technique highlights the utility of converting a complex expression into simpler known limits, effectively employing core trigonometric relationships to find required values.
Other exercises in this chapter
Problem 7
In Problems 7-16, sketch the graph of \(f\). $$ f(x, y)=6 $$
View solution Problem 7
$$ f(x, y)=x y+\frac{2}{x}+\frac{4}{y} $$
View solution Problem 7
In Problems 1-16, find all first partial derivatives of each function. \(f(x, y)=\sqrt{x^{2}-y^{2}}\)
View solution Problem 7
$$ \text { In Problems 1-10, find the gradient } \nabla f \text {. } $$ $$ f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}} $$
View solution