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Math Help - Two variable limits. Continuity and differentiability.

  1. #1
    Newbie dttah's Avatar
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    Two variable limits. Continuity and differentiability.

    Hello everyone, I have the following function:

    \frac{(sen(xy))}{\sqrt{x^2+y^2}} \rightarrow if (x,y) \neq (0,0)

    0 \rightarrow if  (x,y) = (0,0)

    I have to find out its continuity and differentiability in (0,0).

    Concerning the continuity:

    \lim_{(x,y)\rightarrow (0,0)} \frac{(sen(xy))}{\sqrt{x^2+y^2}} \rightarrow \frac{(sen(xy))}{\sqrt{x^2+y^2}} * \frac{xy}{xy} \rightarrow \frac{(sen(xy))}{xy} \frac{xy}{\sqrt{x^2+y^2}}

    senxy/xy goes to 1. While the other one:

    \lim_{(x,y)\rightarrow (0,0)} \frac{xy}{\sqrt{x^2+y^2}}  \leqslant \frac{|xy|}{\sqrt{x^2+y^2}} \leqslant \frac{1}{2} \frac{x^2+y^2}{\sqrt{x^2+y^2}} \Rightarrow \lim_{(x,y)\rightarrow (0,0)} \frac{1}{2}x^2+y^2 = 0

    This is what I did so far, is this correct? If it isn't, why? Concerning the differentiability I'll tell how I did as soon as I understand if I did this right (because if I did this wrong is useless to post the other part!). Thanks a lot!
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  2. #2
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    Re: Two variable limits. Continuity and differentiability.

    Are you sure it's "sen" and not "sin"?
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  3. #3
    Newbie dttah's Avatar
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    Re: Two variable limits. Continuity and differentiability.

    yes, it's sin. I'm sorry I'm italian and the "sinus" is called "seno" aka sen. Anyhow, it's sin!
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    Re: Two variable limits. Continuity and differentiability.

    Quote Originally Posted by dttah View Post
    Hello everyone, I have the following function:

    \frac{(sen(xy))}{\sqrt{x^2+y^2}} \rightarrow if (x,y) \neq (0,0)

    0 \rightarrow if  (x,y) = (0,0)

    I have to find out its continuity and differentiability in (0,0).

    Concerning the continuity:

    \lim_{(x,y)\rightarrow (0,0)} \frac{(sen(xy))}{\sqrt{x^2+y^2}} \rightarrow \frac{(sen(xy))}{\sqrt{x^2+y^2}} * \frac{xy}{xy} \rightarrow \frac{(sen(xy))}{xy} \frac{xy}{\sqrt{x^2+y^2}}

    senxy/xy goes to 1. While the other one:

    \lim_{(x,y)\rightarrow (0,0)} \frac{xy}{\sqrt{x^2+y^2}}  \leqslant \frac{|xy|}{\sqrt{x^2+y^2}} \leqslant \frac{1}{2} \frac{x^2+y^2}{\sqrt{x^2+y^2}} \Rightarrow \lim_{(x,y)\rightarrow (0,0)} \frac{1}{2}x^2+y^2 = 0

    This is what I did so far, is this correct? If it isn't, why? Concerning the differentiability I'll tell how I did as soon as I understand if I did this right (because if I did this wrong is useless to post the other part!). Thanks a lot!
    For a function to be continuous at a point, it needs to be defined at that point, and the limit of the function at that point has to equal the same value. At (x, y) = (0, 0) the function is defined and equals 0. We need to show that the function also approaches that value as you approach (x, y) = (0, 0).

    I agree with how you broke up the limit into \displaystyle \begin{align*} \lim_{(x, y) \to (0, 0)}\frac{\sin{(x\,y)}}{x\,y} \cdot \lim_{(x, y) \to (0, 0)}\frac{x\,y}{\sqrt{x^2 + y^2}} = 1 \cdot \lim_{(x, y) \to (0, 0)}\frac{x\,y}{\sqrt{x^2 + y^2}}\end{align*}

    The easiest way to evaluate this limit will be to convert to polars, remembering that \displaystyle \begin{align*} x = r\cos{\theta}, y = r\sin{\theta} \end{align*} and \displaystyle \begin{align*} x^2 + y^2 = r^2 \end{align*}. Then we note that no matter which path we approach (x, y) = (0, 0) on, the magnitude (r) always approaches 0. So

    \displaystyle \begin{align*} \lim_{(x, y) \to (0, 0)}\frac{x\,y}{\sqrt{x^2 + y^2}} &= \lim_{r \to 0}\frac{r^2\cos{\theta}\sin{\theta}}{r} \\ &= \lim_{r \to 0}r\cos{\theta}\sin{\theta} \\ &= 0 \end{align*}

    So the function is defined to be equal to 0 when (x, y) = (0, 0), and the function also approaches 0 at that point. So the function is continuous at that point.


    As for differentiability, not only does the function need to be continuous at that point, but it also needs continuous partial derivatives at that point. Are you able to evaluate the partial derivatives and see if the partial derivatives are continuous at (0, 0)?
    Thanks from dttah
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  5. #5
    Newbie dttah's Avatar
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    Re: Two variable limits. Continuity and differentiability.

    Hello. First above all I would like to thank you for your help, it was pretty useful.

    Concerning the differentiability, it really takes me a lot to type everything in latex so I will just post an image.

    The first one (cattura1.png) shows how I did the two partial derivatives.
    The second one (cattura.png) I attempted to use the definition of differentiability in a point.

    I am not really sure this is correct.
    Thanks once again for your help.
    Attached Thumbnails Attached Thumbnails Two variable limits. Continuity and differentiability.-cattura1.png   Two variable limits. Continuity and differentiability.-cattura.png  
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