/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Find the limit of \(f\) as \((x,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. \(f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)

Short Answer

Expert verified
The limit does not exist.

Step by step solution

01

Understand the Problem

We need to find the limit of the function \(f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}\) as \((x, y)\) approaches \((0, 0)\). We're going to check if this limit exists or not.
02

Convert to Polar Coordinates

To better analyze the function as \((x, y)\) approaches \((0, 0)\), we can express \(x\) and \(y\) in polar coordinates: \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\), where \(r = \sqrt{x^2 + y^2}\). Then, re-write the function in terms of \(r\) and \(\theta\).
03

Substitute and Simplify

Rewrite the function as follows: \(f(x, y) = \frac{(r\cos(\theta))^2 - (r\sin(\theta))^2}{(r\cos(\theta))^2 + (r\sin(\theta))^2} = \frac{r^2(\cos^2(\theta) - \sin^2(\theta))}{r^2(\cos^2(\theta) + \sin^2(\theta))}\). Notice that the denominator reduces to \(r^2(1)\) because \(\cos^2(\theta) + \sin^2(\theta) = 1\). Therefore, \(f(x, y) = \cos^2(\theta) - \sin^2(\theta) = \cos(2\theta)\).
04

Analyze the Result in Terms of \(\theta\)

The expression \(f(x, y) = \cos(2\theta)\) depends solely on \(\theta\). As \((x, y)\) approaches \((0, 0)\), different paths can correspond to different angles \(\theta\). Since \(\cos(2\theta)\) can vary depending on \(\theta\), the limit is not unique, due to the fact that it changes with \(\theta\).
05

Conclude the Limit Does Not Exist

Since the value of \(\cos(2\theta)\) changes depending on the approach path defined by \(\theta\), there is no single limit value as \((x, y)\) approaches \((0, 0)\). Thus, the limit of the function does not exist.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Polar Coordinates
Polar coordinates offer a way to express the position of a point in terms of a distance and an angle. In our situation, as \(x, y\) approaches \(0, 0\), we convert Cartesian coordinates to polar coordinates using \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). Here, \(r\) is the radial distance from the origin, calculated as \(\sqrt{x^2 + y^2}\), and \(\theta\) is the angle with respect to the positive x-axis.

Converting to polar coordinates helps simplify the limit problem by reducing the dimensionality of the input. Instead of considering movement in two dimensions separately, polar coordinates allow us to view the problem in terms of radial direction and angular orientation. This transformation organizes the problem's behavior as \(r \ ightarrow 0\) and makes it easier to manipulate, often revealing whether or not a limit exists."

In our function, \(f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}\), converting to polar coordinates converts the expression to \(\cos(2\theta)\). This shows that the function does not depend on \(r\), hinting that the approach path's angle \(\theta\) plays a crucial role.
Path Dependence
Path dependence occurs when the result of a function's limit varies depending on the path taken toward a particular point. In multivariable calculus, we often examine limits approaching \(0,0\) by taking various paths such as lines, curves, or even spirals. For our function \(f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}\), substitute \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\), simplifying the expression to \(\cos(2\theta)\). This result makes it evident that the final value is solely dependent on the angle \(\theta\), indicating strong path dependence.

Simply put, two different paths toward \(0,0\) can yield different values for \(f(x,y)\). For example:
  • Along the x-axis: \(y = 0\), then \(\theta = 0\), so \(\cos(2 \times 0) = 1\).
  • Along the y-axis: \(x = 0\), then \(\theta = \frac{\pi}{2}\), so \(\cos(2 \times \frac{\pi}{2}) = -1\).

Since varying paths lead to different values, arriving at a consistent limit value is impossible. This path dependence shows why the function does not have a well-defined limit.
Limit Does Not Exist
When we say "the limit does not exist," it means there isn't a single value the function approaches as the input gets arbitrarily close to a particular point. The limit should be unique regardless of the approach path. For the given function \(f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}\), analyzing the expression in polar coordinates \(\cos(2\theta)\) shows that different paths yield different limiting values.

The central issue here is that each approach direction, corresponding to a unique angle \(\theta\), results in a different output for \(\cos(2\theta)\). Therefore, there is no single value the function converges to, making the limit undefined. The path dependency finding from our exploration into polar coordinates supports the conclusion that
  • If a single limit existed, all paths reaching \(0,0\) would provide the same function value.
  • Since they do not, the limit cannot be defined consistently.

So, for functions like these, if changing the path changes the result, the limit does not exist. It's a key finding about the behavior of multivariable functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises \(49-54\) , use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Maximize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(2 y+4 z-5=0\) and \(4 x^{2}+4 y^{2}-z^{2}=0\)

Find $$ \text { a. }\left(\frac{\partial w}{\partial y}\right)_{x} \quad \text { b. }\left(\frac{\partial w}{\partial y}\right)_{z} $$ at the point \((w, x, y, z)=(4,2,1,-1)\) if $$ w=x^{2} y^{2}+y z-z^{3} \quad \text { and } \quad x^{2}+y^{2}+z^{2}=6 $$

Express \(v_{x}\) in terms of \(u\) and \(y\) if the equations \(x=v \ln u\) and \(y=u \ln v\) define \(u\) and \(v\) as functions of the independent variables \(x\) and \(y,\) and if \(v_{x}\) exists. (Hint: Differentiate both equations with respect to \(x\) and solve for \(v_{x}\) by eliminating \(u_{x}\).)

Least squares and regression lines When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$ \begin{array}{l}{m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}}} \\\ {b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right)}\end{array} $$ with all sums running from \(k=1\) to \(k=n .\) Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\) 3\. handle data analytically.

a. Around the point \((1,0),\) is \(f(x, y)=x^{2}(y+1)\) more sensitive to changes in \(x\) or to changes in \(y ?\) Give reasons for your answer. b. What ratio of \(d x\) to \(d y\) will make \(d f\) equal zero at \((1,0) ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.