Study the continuity and differentiability of function in point $(0,0) $ where $f:\mathbb{R}^2\rightarrow \mathbb{R}$

$f(x,y)= \begin{cases} 0 & \left(x,y\right)\:=\left(0,\:0\right) \\ \frac{x^3y}{x^4+y^2} & \left(x,y\right)\ne \left(0,0\right) \\ \end{cases} $

I can see that the function will go towards 0 as (x,y) goes towards 0, so that makes it continuous. But what about differentiability?

The definition I got in class is rather fuzzy. Basically, it's like this: Step 1. Check if the function can be derived in point a - in my case $(0,0)$ If it isn't the function wont be differentiable in point a. If it is, calculate the partial derivatives and proceed to step 2.

Step 2. Study the limit: $$l=\lim _{h->O_m}\left(\frac{f\left(a+h\right)-f\left(a\right)-<h,\Delta f\left(a\right)>}{\left|\left|h\right|\right|}\right)$$ if $l=O_m$ then the function is differentiable in point a. Otherwise it is not.

Also, $l=\lim _{\left(h_1,....,h_m\right)\rightarrow O_m}\left(\frac{f\left(a_1+h_1+...+a_m+h_m\right)-f\left(a_1,...,a_m\right)-\sum h_j\left(\frac{\:∂L}{\:∂x_j}\left(\right)\right)}{\sqrt{h_1^2+...+h_m^2}}\right)$

My problem is that I don't really understand the definition. I also might've written it down wrong, and I also have problems understand the terms that appear in it. I tried looking online but what I could find didn't really answer all my questions. Can someone help me understand this, so that I can solve exercises such as these, and possibly with more variables as well, in other points?

  • $\begingroup$ $(x,y)$ doesn't go to $0,$ it goes to $(0,0).$ $\endgroup$ – zhw. Jun 21 '16 at 23:48
  • $\begingroup$ "The definition I got in class is rather fuzzy." That's not a definition, it's a recipe. $\endgroup$ – zhw. Jun 21 '16 at 23:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.