1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Differentiability of a two variable function with parameter

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known data

    For wich parameter [tex]\alpha\in\mathbb{R}[/tex] the function:
    [tex]f(x, y)= \begin{cases}|x|^\alpha \sin(y),&\mbox{ if } x\ne 0;\\ 0, & \mbox{ if } x=0\end{cases}[/tex]

    is differentiable at the point (0, 0)?

    3. The attempt at a solution

    For α<0, the function is not continue at (0, 0), so it is not differentiable. I checked it :)

    For α≥0 i have troubles, many troubles. I need your help.

    I started evalueting the partial derivates using the definition:

    [tex]\partial_x f(0,0):= \lim_{h\to 0} \frac{|h|^\alpha\sin(0)-0}{h}=0 [/tex]
    [tex]\partial_y f(0,0):= \lim_{h\to 0} \frac{|0|^\alpha\sin(h)-0}{h}=0 [/tex]

    (I think i must split the cases α=0 and α>0, right? |0|^α has no sense if α=0, but if α=0 then f(x,y)= sin(y)... I'm confuse)

    Anyway, i need to find the value of the limit (if it exists):

    [tex]\lim_{(h,k)\to (0, 0)}\frac{f(h, k)}{\sqrt{h^2+k^2}}[/tex] and show that its value is 0 right? I have problems with two variables limits...

    Any helps will be appreciated...

    Please, if you see there are mistakes in English languange, correct me :)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 9, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    See if the limit of each of the following exists as (x, y) → (0, 0):
    [itex]\displaystyle \frac{\partial f(x,\,y)}{\partial x}[/itex] and [itex]\displaystyle \frac{\partial f(x,\,y)}{\partial y}[/itex]​
    Last edited: Nov 9, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook