Yes but then how are we going to evaluate this limit?

I mean if

$\displaystyle

\lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}

$

And in the expression on the RHS there are no more $\displaystyle \Delta_x, \Delta_y$ terms left, and we don't know the value of m! So, what do I need to do?

You need to really concentrate a little more: if you put $\displaystyle \Delta_y=m\Delta_x$ , then $\displaystyle (\Delta_x,\Delta_y)\rightarrow (0,0) \Longleftrightarrow \Delta_x\rightarrow 0$ , so THE LIMIT IS $\displaystyle \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}$ , and then the limit depends on $\displaystyle m$ (unless you could prove the limit is the same no matter what m we choose...which of course you won't be able to prove since it isn't true ). Tonio
Of course, I could rationalize the denominator but I don't know how to simplify the numerator in the following:

$\displaystyle

\lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}

$