Show That Y x x2 is Continuous Everywhere
Prove $f(x,y) = xy/(x^2 + y^2)$ is continuous everywhere except $(0,0).$
In a simplified and quicker approach, just consider those points where $f$ is not well defined, to identify non-continuity. You need more care in your discussion on " $h$ is not defined". In this case, the only point which gives "trouble" for $f$ is $(0,0)$. But we must check this formally! Using polar coordinates $(\rho,\theta)$ you can prove that
$$\lim_{(x,y)\rightarrow (0,0)}f(x,y) ~~(*)$$
is equal to the limit
$$\lim_{\rho\rightarrow 0}\cos\theta\sin\theta:=q(\theta), $$
which is a function of the angle $\theta$: in summary the limit $(*)$ depends on the path chosen to approach $(0,0)$, which means that $f$ is not continuous at $(0,0)$.
For all points different from $(0,0)$ just use the definition of continuity of a function $f:\mathbb R^2\rightarrow \mathbb R$ (with $\epsilon-\delta$, for example); the thesis follows.
Related videos on Youtube
03 : 44
Find the values a and b that make the piecewise function continuous
11 : 48
Show that f(x,y) = xy/√(x2+y2) is continuous at origin when f(0,0)=0
03 : 06
Multivariable Calculus Limit of y^2/(x^2 + y^2) as (x,y) approaches (0,0)
11 : 20
Continuity of Multivariable Functions
09 : 22
Proof: f(x) = sqrt(x) is Continuous using Epsilon Delta Definition | Real Analysis
Comments
-
I'd just like to ask you if my proof here is valid. I'll provide you with the method I used and if it seems ok let me know! If not, explanations would be helpful!
My main approach to this question involves using the theorem that if functions $g$ and $h$ are both continuous, then their product and their sum are also continuous.
I begun by splitting $f$ into $(g)(1/h)$, where $g(x,y) = xy$ and $h(x,y) = x^2 + y^2.$
Then, I again split $g$ into $2$ functions $p(x) = x$ and $q(x) = y$ and claimed that since they were both polynomials of order 1, they must be continuous everywhere. (Is it ok to just say this? It's a theorem in my book.) Hence, their product $xy$ must be continuous.
Similarly, the sum of $x^2$ and $y^2$ (two polynomials of order $2$) must also be continuous.
Now, we are left with $f(x,y) =$ the product of $g$ and the inverse of $h$. The inverse of $h$ is again continuous everywhere, except $(0,0)$ as $h(0,0)$ is undefined. Thus, $f(x,y)$ is continuous everywhere, except $(0,0)$.
Thanks for reading!
-
The inverse of $h(0,0)$ is undefined, $h(0,0)$ is just equal to $0$.
-
Last paragraph needs more investigation; it is true that $1/h$ is undefined at $(0,0)$, but $f$ is the product of $g$ and $1/h$. In other words, you need to check continuity at $(0,0)$ at the whole product $g/h$. Please look at my answer.
-
As Mr. Avitus said, It is necessary to check the behaviour of both $g$ and $h$ combined at any point...
-
@Alan Brett : you do not need to think about the limit of $f(x,y)$ as $(x,y) \to (0,0)$ to answer your question. $f$ cannot be continuous at $(0,0)$ since $f(0,0)$ doesn't exist. If you had assigned a value to $f(0,0)$ you would have to do it, but you didn't.
-
Ahhhh,the prototypical limit problem of functions of 2 variables in calculus. How many of us have seen this problem in our formative days? It IS a really good problem because this really has all the earmark properties of a discontinuity in higher dimensions.
-
-
I have tried a similar method to yours just now! I approached (0,0) along lines of form 'y = mx'. Substituted that into f(x,y), getting f(x,mx) = m/(1+m^2). Then I showed that if you approach along y=x, the limit = 1/2. However, approaching from y= -x, the limit = -1/2. Hence, this limit does not exist.
-
exactly!Well done.
-
Technically the function has not been defined at the origin, so even if it had a limit, I'm not sure one should say it is continuous there, only that it is possible to extend it continuously.
Recents
Related
Source: https://9to5science.com/prove-f-x-y-xy-x-2-y-2-is-continuous-everywhere-except-0-0
0 Response to "Show That Y x x2 is Continuous Everywhere"
Post a Comment