I need to prove that $\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$ I need to prove the following limit
$$\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$$
Using the Squeeze Theorem 
$$0\le|\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}|\leq\frac{|x|\log(1+y)}{y}\to0$$
For $(x,y)\to(0,0)$
Here I have used the fact that 
$$\sqrt{x^2+y^2}\ge\sqrt{y^2}=y$$
Such that
$$\frac{1}{\sqrt{x^2+y^2}}\le\frac{1}{y}$$
And the fact that
$$\lim_{y\to0}\frac{\log(1+y)}{y} =0$$
So I can conclude that
$$0\le\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}\le\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{y}\le0$$
And thus by the Squeeze Theorem 
$$\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$$
Is my proof correct? 
 A: Almost correct. Except  $\lim_{y\to0}\frac{\log(1+y)}{y} = 1 \neq 0$.
Here's how  I would  do  it:
Because $(|x|-|y|)^2\geq 0$, we have $x^2+y^2\geq 2|xy|$ (this  is also known as AM-GM inequality).
Therefore, $$\frac 1 {\sqrt{x^2+y^2}}\leq \frac 1 {\sqrt{2|xy|}}$$
and $$\left|\frac{x\log(1+y)}{\sqrt{x^2+y^2}}\right|\leq \frac{\sqrt{|x|}}{\sqrt 2} \cdot\frac{|\log(1+y)|}{\sqrt{|y|}}$$
Now $$\lim_{y\rightarrow 0}\frac{|\log(1+y)|}{\sqrt{|y|}}=\lim_{y\rightarrow 0}\frac{|\log(1+y)|}{|y|}\sqrt{|y|}=1\cdot 0 = 0$$
So by the squeeze theorem, you  get what you want.
A: Carefull: $\sqrt{y^2}=|y|$ and
$$\lim_{y\to0}\frac{\log(1+y)}{y} =\lim_{y\to0}\log(1+y)^{1\over y}= \log e =1$$
A: $\begin{aligned}
& \lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}} \xrightarrow{\begin{cases}x=r \cdot \cos\theta
\\ y=r \cdot \sin\theta
\end{cases}}\lim_{r\to 0}{\frac{|r \cdot \cos\theta| \ln{(1+r \cdot \sin\theta)} }{ \sqrt{(r \cdot \cos\theta)^2+(r \cdot \sin\theta)^2} } }
\\& \le \lim_{r\to 0}\frac{|r \cdot \cos\theta|(r\cdot \sin\theta)}{ \sqrt{(r \cdot \cos\theta)^2+(r \cdot \sin\theta)^2} }= \lim_{r\to 0}\frac{r ^2\cdot| \cos\theta|(\sin\theta)}{r}=\lim_{r\to 0} r\cdot| \cos\theta|(\sin\theta)
\\& \le \lim_{r\to 0} r\cdot1 =0
\end{aligned}$
