Question: I came across the following limit 

$$\lim\_{(x,y)\to(0,0)} \frac{y \sqrt{x}}{\sqrt{x^2 + y^2}}$$

If $x\geq0$ then 

$$0 \leq \bigg| \frac{y \sqrt{x}}{\sqrt{x^2 + y^2}}\bigg|= \frac{|y| \sqrt{x}}{\sqrt{x^2 + y^2}} \leq \sqrt{x}$$

So using the squeeze theorem the limit is equal to zero.

However for any neighborhood $V$ of $(0,0)$ there are points where $x \lt 0$ and therefore $f$ is not defined . Does it mean that the limit doesn't exist, or should we only consider the points where $f$ is defined?

Answer: 
If we know that, then\begin{align}\exp'(x)&=\left(\sum\_{n=0}^\infty\frac{x^n}{n!}\right)'\\&=\sum\_{n=0}^\infty\frac{x^n}{n!}\\&=\exp(x).\end{align}Therefore,\begin{align}\ln'(x)&=\frac1{\exp'(\ln(x))}\\&=\frac1{\exp(\ln(x))}\\&=\frac1x.\end{align}
Let $f(x)=e^x$. Then:

$${\partial \log (x) \over \partial x} = {\partial f^{-1}(x) \over \partial x} = {1 \over e^{\log(x)}} = 1/x$$.