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: 
Let $f(x)=e^x$, so $f^{-1}(x)=\ln x$. We know that $(f \circ f^{-1})(x)=x$. Differentiate by the chain rule to get
$$
f'(f^{-1}(x)) \frac{d}{dx}f^{-1}(x)=1.
$$
As $f'=f$ this boils down to
$$
x \frac{d}{dx}f^{-1}(x)=1
$$
giving what you want.
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}