# 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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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: $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^4+y^2}=\lim_{r\rightarrow 0}\frac{r^3\cos(\theta)^2\sin(\theta)}{r^4\cos(\theta)^4+r^2\sin(\theta)^2}=0$$
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?