- #1
twoflower
- 363
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Hello to everybody.
Yesterday we wrote a fake test, in which I encountered problem I haven't benn able to solve so far.
The problem is given this way:
Let
<br /> f(x,y) := \left\{\begin{array}{cc}\frac{xy^3}{x^2+y^4} - 2x + 3y,&[x,y] \neq [0,0]\\0, &[x,y] = [0,0]\end{array}\right.<br />
Find out, whether the function f[/tex] in point [0,0] is continuous<br /> <br /> Well, to be continuous in origin, it has to have zero limit as [x,y] approaches [0,0].<br /> <br /> I get<br /> <br /> <br /> \lim_{[x,y] \rightarrow [0,0]} \frac{xy^3}{x^2+y^4} - 2x + 3y<br /><br /> <br /> after converting to polar coordinates (which proved to be very useful when limiting two-variable functions) I have<br /> <br /> <br /> \lim_{r \rightarrow 0} \frac{r^4\cos \varphi \sin^3 \varphi}{r^2\cos^2 \varphi\ +\ r^4\sin^4 \varphi}\ -\ 2\lim_{r \rightarrow 0} r\cos \varphi\ +\ 3\lim_{r \rightarrow 0} r\sin \varphi = \lim_{r \rightarrow 0} r\cos\varphi \sin^3 \varphi\ \frac{r}{\cos^2 \varphi\ +\ r^2 \sin^4 \varphi}<br /><br /> <br /> The first part goes to zero so it would be sufficient to show that the second part is bounded. And that's where I got stuck. <br /> <br /> Horse sense tells me this: if \varphi is such that \varphi is non-zero, the whole limit is zero. If \cos \varphi is zero, then the limit is still zero...but I'm not sure of this and definitely it's not an acceptable mathematical solution. But I'm unable to prove this elegantly at this point...<br /> <br /> Thank you very much for pointing me to the right direction.
Yesterday we wrote a fake test, in which I encountered problem I haven't benn able to solve so far.
The problem is given this way:
Let
<br /> f(x,y) := \left\{\begin{array}{cc}\frac{xy^3}{x^2+y^4} - 2x + 3y,&[x,y] \neq [0,0]\\0, &[x,y] = [0,0]\end{array}\right.<br />
Find out, whether the function f[/tex] in point [0,0] is continuous<br /> <br /> Well, to be continuous in origin, it has to have zero limit as [x,y] approaches [0,0].<br /> <br /> I get<br /> <br /> <br /> \lim_{[x,y] \rightarrow [0,0]} \frac{xy^3}{x^2+y^4} - 2x + 3y<br /><br /> <br /> after converting to polar coordinates (which proved to be very useful when limiting two-variable functions) I have<br /> <br /> <br /> \lim_{r \rightarrow 0} \frac{r^4\cos \varphi \sin^3 \varphi}{r^2\cos^2 \varphi\ +\ r^4\sin^4 \varphi}\ -\ 2\lim_{r \rightarrow 0} r\cos \varphi\ +\ 3\lim_{r \rightarrow 0} r\sin \varphi = \lim_{r \rightarrow 0} r\cos\varphi \sin^3 \varphi\ \frac{r}{\cos^2 \varphi\ +\ r^2 \sin^4 \varphi}<br /><br /> <br /> The first part goes to zero so it would be sufficient to show that the second part is bounded. And that's where I got stuck. <br /> <br /> Horse sense tells me this: if \varphi is such that \varphi is non-zero, the whole limit is zero. If \cos \varphi is zero, then the limit is still zero...but I'm not sure of this and definitely it's not an acceptable mathematical solution. But I'm unable to prove this elegantly at this point...<br /> <br /> Thank you very much for pointing me to the right direction.
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