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Recently I came upon a limit which confused me. The reason is that when I try to solve the following limit using polar coordinates I get a constant which I do not know if it gives me information.

Let : $$\lim_{x,y \to 0} \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}$$ Using polar coordnates I get this: $$\lim_{r \to 0} \frac{r\cos\left(\theta\right)+r\sin\left(\theta\right)-\frac{1}{2}r^2\sin^2\left(\theta\right)}{\sin\left(r\sin\left(\theta\right)\right)+\log\left(1+r\cos\left(\theta\right)\right)}$$

Which is equal to: $$\lim_{r \to 0} \frac{r\cos\left(\theta\right)+r\sin\left(\theta\right)-\frac{1}{2}r^2\sin^2\left(\theta\right)}{r\sin\left(\theta\right)+r\cos\left(\theta\right)}=1$$ I already know this limit does not exist. Actually it was quite difficult to find a path for which I get a different limit...

My question is: If I use polar coordinates and the result is not something that depends on $r,\theta$ then what I get is basically useless information? (I know that if that limit goes to infinity the limit of the function does not exist)

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  • $\begingroup$ I have doubts about going from the second equation to the third. $\endgroup$
    – MPW
    May 29, 2018 at 22:56
  • $\begingroup$ I have been fighting with this limit since yesterday, may be I made a mistake. But I can not see it right now. $\endgroup$
    – TheNicouU
    May 29, 2018 at 22:58
  • $\begingroup$ My first thought would be to consider paths of approach where the numerator is constant. Then think about the cases with the constant being $0$ and it being $1$. If you get a nonzero limit in the second case, you are done since the limit in the first case is zero. $\endgroup$
    – MPW
    May 29, 2018 at 22:59
  • $\begingroup$ Actually if you approach with the parabola $x=y^2-y$ the limit is equal to 1/2 which proves the limit does not exist. $\endgroup$
    – TheNicouU
    May 29, 2018 at 23:02
  • $\begingroup$ @TheNicouU Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Jun 22, 2018 at 21:04

1 Answer 1

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Note that

  • $x=0,\, y=t\to 0 \implies \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}=\frac{t-\frac{1}{2}t^2}{\sin\left(t\right)}\to 1$

  • $x=-t+\frac12t^2,\, y=t,\, t\to 0 \implies \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)} =\frac{-t+\frac12t^2+t-\frac12t^2}{\sin\left(t\right)+\log\left(1-t+\frac12t^2\right)}=0$

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