- #1
twoflower
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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
[tex] 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.[/tex]
Find out, whether the function [itex]f[/tex] in point [0,0] is continuous[/itex][itex] <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 /> [tex] \lim_{[x,y] \rightarrow [0,0]} \frac{xy^3}{x^2+y^4} - 2x + 3y[/tex]<br /> <br /> after converting to polar coordinates (which proved to be very useful when limiting two-variable functions) I have<br /> <br /> [tex] \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}[/tex]<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 [itex]\varphi[/itex] is such that [itex]\varphi[/itex] is non-zero, the whole limit is zero. If [itex]\cos \varphi[/itex] 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.[/itex]
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
[tex] 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.[/tex]
Find out, whether the function [itex]f[/tex] in point [0,0] is continuous[/itex][itex] <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 /> [tex] \lim_{[x,y] \rightarrow [0,0]} \frac{xy^3}{x^2+y^4} - 2x + 3y[/tex]<br /> <br /> after converting to polar coordinates (which proved to be very useful when limiting two-variable functions) I have<br /> <br /> [tex] \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}[/tex]<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 [itex]\varphi[/itex] is such that [itex]\varphi[/itex] is non-zero, the whole limit is zero. If [itex]\cos \varphi[/itex] 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.[/itex]
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