I'm asked to calculate this limit: $$ \lim_{(x,y)\rightarrow(0,0)}\frac{\ln(1+x)+\ln(1+y)}{x+y} $$
After calculating iterated limits and using some directions ($y=\lambda x$ and $ y=\lambda x^2$) all I can deduce is that the limit is actually $1$, but using polar coordinates I'm not able to prove it. Is there any other way to solve it?
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Consider the path $y=\frac1{1+x}-1$: $$ \lim_{x\rightarrow 0}\frac{\ln(1+x)+\ln\left(1+\frac1{1+x}-1\right)}{x+\frac1{1+x}-1}= \lim_{x\rightarrow 0}\frac{\ln(1+x)-\ln(1+x)}{x+\frac1{1+x}-1}=0. $$
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Wrong answer. I did not delete for I was asking things in the comments.
If you're allowed to use Taylor series, then here is another way: consider
$$\ln(1+x) \sim x - \dfrac{x^2}{2} + O(x^3)$$
and the same for $\ln(1+y)$, clearly when $x\to 0$ and $y\to 0$.
In this way we have:
$$\lim_{(x, y)\to (0,0)} \dfrac{x - \dfrac{x^2}{2} + y - \dfrac{y^2}{2}}{x+y} = \dfrac{x+y - \dfrac{1}{2}(x^2+y^2)}{x+y} = 1 - \dfrac{1}{2}\dfrac{(x+y)^2- 2xy}{x+y}$$
Now
$$ \dfrac{1}{2}\dfrac{(x+y)^2- 2xy}{x+y} = \dfrac{1}{2}\left[ x+y - \dfrac{2xy}{x+y}\right]$$
Since $x\to 0$ and $y\to 0$ we can think of them as two infinitesimals (say $\epsilon$) hence
$$\dfrac{2xy}{x+y} \sim \dfrac{2\epsilon^2}{\epsilon} = 2\epsilon \to 0$$
Whence your limit is $1$.
Choose $\alpha \neq 0$ and choose the following path $s=x+\alpha x^2, t=-x+\alpha x^2$. \begin{eqnarray} {\log (1+s)+\log(1+t) \over s+t} &=& { \log ( (1+\alpha x^2)^2 - x^2) \over 2 \alpha x^2 } \\ &=& { \log ( (1+\alpha x^2)^2 ( 1 - {x^2 \over (1+\alpha x^2)^2 } ) ) \over 2 \alpha x^2 } \\ &=& { 2\log (1+\alpha x^2) \over 2 \alpha x^2 } + { \log ( 1 - {x^2 \over (1+\alpha x^2)^2 } ) \over 2 \alpha x^2 } \end{eqnarray} A little L'Hôpital shows that the limit as $x \to 0$ is $1-{1 \over 2 \alpha}$.
In particular, the limit in the question does not exist.