# Solve $\lim_{(x,y) \to (0,0)} \frac{\ln(1+xy)}{\sqrt{x^2+y^2}}$ and $\lim_{(x,y) \to (0,0)} \frac{x^2 y}{x^4+y^2}$

I am interested to find the limit of the following functions,

$$\lim_{(x,y) \to (0,0)} \frac{\ln(1+xy)}{\sqrt{x^2+y^2}}$$

$$\lim_{(x,y) \to (0,0)} \frac{x^2 y}{x^4+y^2}$$

If the limit is not existing then what should be the reason of its not existance.

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There is another post about the second limit: math.stackexchange.com/questions/1476353/… – Martin Sleziak Oct 12 '15 at 14:18
Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Oct 12 '15 at 14:32

## 2 Answers

By using the polar transformation, we have: $$\lim_{(x,y)\rightarrow(0,0)}\frac{\ln(1+xy)}{\sqrt{x^2+y^2}}=\lim_{r\rightarrow 0}\frac{\ln(1+r^2\sin(\theta)\cos(\theta))}{r}=\lim_{r\rightarrow 0}\frac{r^2\sin(\theta)\cos(\theta)}{r}=0$$ Also: $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^4+y^2}=\lim_{r\rightarrow 0}\frac{r^3\cos(\theta)^2\sin(\theta)}{r^4\cos(\theta)^4+r^2\sin(\theta)^2}=0$$

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I do not understand how you removed the ln in the last step before reaching to 0? – Kiran Mar 10 '14 at 14:55
For $x\rightarrow0$, $ln(1+x)$ and $x$ are equivalent infinitesimal. – Lion Mar 10 '14 at 15:09

Along the path $y=x^2$ shouldn't the second limit equal $\frac{1}{2}$? Since along the x-axis or the y-axis the limit equals zero, wouldn't that mean that the second limit doesn't exist?

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You are quite correct. (And welcome to math.stackexchange!) – mrf Sep 18 '15 at 14:35