<s>[INST] 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 $x,y = {2\over n},{1 \over n}$
 [/INST] Substitute $x= r \cos \theta$ and $y= r \sin \theta$, then you have
$$\lim_{r \to 0}\frac{r \cos \theta \ln(1+r \sin \theta)}{2 r^2 \cos ^2 \theta + r^2 \sin^2 \theta}=\lim_{r \to 0}\frac{ \cos \theta \ln(1+r \sin \theta)}{2 r \cos ^2 \theta + r \sin^2 \theta}\stackrel{l'Hospital}{=}\lim_{r \to 0}\frac{\frac{\cos \theta \sin\theta}{1+r \sin \theta}}{2 \cos^2\theta + \sin^2 \theta}=\lim_{r \to 0}\frac{\cos \theta\sin \theta}{(2 \cos^2 \theta + \sin^2\theta)(1+r \sin \theta)}=\frac{\cos \theta\sin \theta}{2\cos^2 \theta+\sin^2 \theta}.$$
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