Finding $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$ Find bounds for $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$


I am finding maximum and minimum for function and one of critical case is to find possible minimal and maximal value of given function in $0,0$. But how can I do this due to this limit doesn't exists (for example we can take $x,y = {1\over n},{1\over n}$ and $x,y = {2\over n},{1 \over n}$
Every level curve 

$$ f(x,y)= \frac{x \ln(1+y)}{2x^2+y^2}=c $$

for $c\le0.353$ passes through $(0,0)$ but not so for $c&gt;0.354$, so the maximal point $(0,0,z_{\text{MAX}})$ should have $z_{\text{MAX}}$ lie somewhere between those two quantities.

For each level curve, the point nearest $(0,0)$ appears to lie along the line $y=\sqrt{2}x$. Replacing $y$ with $\sqrt{2}x$ gives

$$ y=\frac{\ln(1+\sqrt{2}x)}{4x} $$

and

$$ \lim_{x\to0}\frac{\ln(1+\sqrt{2}x)}{4x}=\frac{\sqrt{2}}{4}\approx0.3535533906 $$

This would be the correct maximum if indeed the point on each level curve closest to $(0,0)$ actually lies along the line $y=\sqrt{2}x$.
Let's have $f(x,y)=\dfrac{x\ln(1+y)}{2x^2+y^2}$

For $x\neq 0$ we have $f(x,0)=0\to 0$

On the other hand $f(x,x)=\dfrac{x\ln(1+x)}{3x^2}=\dfrac{\ln(1+x)}{3x}\sim\dfrac x{3x}\to \dfrac 13$

Since $f$ has at least two different limits along different paths, then $f$ has no limit in $(0,0)$