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Chapter 19: Problem 13

In Exercises 13 through 16, prove that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) exists. \(f(x, y)=\frac{x y}{\sqrt{x^{2}+y^{2}}}\)

Short Answer

Expert verified
The limit exists and is 0.

Step by step solution

01

- Approach the origin along the x-axis

Evaluate the limit by setting y = 0: \[ f(x, 0) = \frac{x \times 0}{\sqrt{x^2 + 0}} = 0 \] So, \( \lim_{x \to 0} f(x,0) = 0 \).
02

- Approach the origin along the y-axis

Evaluate the limit by setting x = 0: \[ f(0, y) = \frac{0 \times y}{\sqrt{0 + y^2}} = 0 \] So, \( \lim_{y \to 0} f(0,y) = 0 \).
03

- Approach along the line y = x

Evaluate the limit by setting y = x: \[ f(x, x) = \frac{x \times x}{\sqrt{x^2 + x^2}} = \frac{x^2}{\sqrt{2x^2}} = \frac{x^2}{x \sqrt{2}} = \frac{x}{\sqrt{2}} \] As \( x \to 0 \), \( \frac{x}{\sqrt{2}} \to 0 \).
04

- Approach along the line y = kx

Generalize the approach by setting y = kx where k is a constant: \[ f(x, kx) = \frac{x(kx)}{\sqrt{x^2 + (kx)^2}} = \frac{kx^2}{\sqrt{x^2 (1 + k^2)}} = \frac{kx^2}{x\sqrt{1 + k^2}} = \frac{kx}{\sqrt{1 + k^2}} \] As \( x \to 0 \), \( \frac{kx}{\sqrt{1 + k^2}} \to 0 \).
05

- Conclude the limit

Since \( f(x, y) \) approaches 0 along different paths, we can conclude that: \[ \lim_{(x, y) \to (0, 0)} \frac{x y}{\sqrt{x^2 + y^2}} = 0 \] Thus, the limit exists and is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Approaching the Origin
In multivariable calculus, understanding how a function behaves as variables approach a certain point is crucial. When finding the limit as \( (x, y) \to (0,0) \), we need to consider how the function behaves as both \( x \) and \( y \) get closer to 0. One common point to study is the origin because it often reveals interesting properties about the function. We examine the function along various paths leading to the origin to ensure that the limit is consistent no matter the direction of approach.
This step-by-step process helps us grasp the overall behavior of the function near the point of interest.
Limits Along Different Paths
When dealing with multivariable limits, it's essential to evaluate the limit along different paths. The function \( f(x, y) = \frac{x y}{\frac{x^2 + y^2}} \) must be approached in several ways to ensure the limit exists.
  • First, we approach along the x-axis by setting y = 0, simplifying the function to \( \frac{x \times 0}{\frac{x^2}} = 0 \).
  • Second, we approach along the y-axis by setting x = 0, simplifying to \( \frac{0 \times y}{\frac{y^2}} = 0 \).
  • We also approach along the line y = x which gives us \( \frac{x \times x}{\frac{2x^2}} = \frac{x}{\frac{2}} \), leading to 0 as x goes to 0.
  • Finally, to generalize, we approach along y = kx where k is any constant. Simplifying this results in the expression \( \frac{kx}{\frac{1 + k^2}} = 0 \).
Each path leads us to the same limit, confirming that our function approaches 0 no matter the path taken.
Proving Limits
To conclusively prove that a limit exists in multivariable calculus, showing the same limit value along multiple paths is paramount. In our exercise, after considering several different approaches, we find consistently that \( \frac{x y}{\frac{x^2 + y^2}} \) approaches 0. Since all paths give us 0, we can soundly conclude that:
\[\frac{x y}{\frac{x^2 + y^2}} = \frac{x y}{x^2 + y^2}\to 0 \tex\frac{(x, y) \to (0, 0)}}\] \ In simpler terms, showing this convergence along all paths verifies that the function behaves uniformly as it approaches the origin. Therefore, the limit exists, ensuring a deeper understanding of the function's properties.

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Most popular questions from this chapter

In Exercises 5 through 8 , prove that \(f\) is differentiable at all points in its domain by doing each of the following: (a) Find \(\Delta f\left(x_{0}, y_{0}\right)\) for the given function; (b) find an \(\epsilon_{1}\) and an \(\epsilon_{2}\) so that Eq. (3) holds; (c) show that the \(\epsilon_{1}\) and the \(\epsilon_{2}\) found in part (b) both approach zero as \((\Delta x, \Delta y) \rightarrow(0,0)\). $$ f(x, y)=\frac{x^{2}}{y} $$

In Exercises 13 through 24 , find the indicated partial derivatives by holding all but one of the variables constant and applying theorems for ordinary differentiation. $$ u=\tan ^{-1}(x y z w) ; \frac{\partial u}{\partial w} $$

In Exercises 1 through 6 , discuss the continuity of \(f\). \(f(x, y)= \begin{cases}\frac{x^{2} y^{2}}{\left|x^{3}\right|+\left|y^{3}\right|} & \text { if }(x, y) \neq(0,0) \\\ 0 & \text { if }(x, y)=(0,0)\end{cases}\)

If \(f\) and \(g\) are differentiable functions of \(x\) and \(y\) and \(u=f(x, y)\) and \(v=g(x, y)\), such that \(\partial u / \partial x=\partial v / \partial y\) and \(\partial u / \partial y=\) \(-\partial v / \partial x\), then if \(x=r \cos \theta\) and \(y=r \sin \theta\), show that $$ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \text { and } \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} $$

Given \(u=9 x^{2}+4 y^{2}, x=r \cos \theta, y=r \sin \theta\). Find \(\partial^{2} u / \partial r^{2}\) in three ways: (a) by first expressing \(u\) in terms of \(r\) and \(\theta\); (b) by using the formula of Example 5; (c) by using the chain rule.
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