Can I use l'Hôpital to find a limit after changing to polar coordinates? I was wondering if there is another approach instead of using the $log (x + 1) \sim_{0} x$ or Taylor series to solve:
$$\lim_{(x,y) \to (0,0)} f(x,y)={\ln(xy + 1) \over x}.$$
Particularly, I'm wondering whether I can use polar coordinates and l'Hopital.
Something like this. First, we switch to polar coordinates:
$$\lim_{(r) \to (0)} f(r)={\ln(r^2\cos\phi \sin\phi + 1) \over r\cos \phi}.$$
Now we use l'Hopital:
\begin{align}
&\lim_{(r) \to (0)} f(r)=\lim_{(r) \to (0)}{\ln(r^2\cos\phi \sin\phi + 1) \over r\cos \phi}
\\
=&\lim_{(r) \to (0)}{{d\over dr}(\ln(r^2\cos\phi \sin\phi + 1) \over{d\over dr} (r\cos \phi)}
\\
=&\lim_{(r) \to (0)}{2r\sin\phi \cos\phi \over {r^2\sin\phi \cos^2\phi + \cos\phi}}
\\
=&\lim_{(r) \to (0)}{2r\sin\phi \over r^2\sin\phi \cos\phi +1}=0.
\end{align}
As $$\lim_{(r) \to (0)} f(\phi)= {2r\sin\phi }= 0.$$
Does this make any sense at all or I can't use l'hopital in this case? 
 A: In L'Hospital rule it was assumed that the limit is of the form $\frac{0}{0}$ on the limit and in the latter case this isn't true. There is also another problem with calculating the limit radially. Consider the function $f(x,y) = \Big\{\begin{eqnarray} 1, & \ y = x^2 \\ 0 ,  & \ y \neq x^2 \end{eqnarray}$.
This has the property $\lim_{r \rightarrow 0} f(r,\theta) = 0$ for every $\theta$, but $f$ has no limit at origo. Now recall the definition of limit at $x_0$. For every $\epsilon > 0$ there is $\delta > 0$ s.t. $|x-x_0| < \delta$ implies $|f(x)-a| < \epsilon$. That means that you have to calculate an estimate for finite neigborhoods of $x_0$ and then show that those estimates go to $0$. L'hospital rule isn't good for that purpose because it gives only the limit without knowledge of the estimates.
A: A few issues:
Why are you using $r$ in some places and $p$ in others?
You are missing a factor of $r$ in your substitution of $x y$.
You should not be applying L'Hopital's rule a second time as your numerator and denominator do not both tend to zero or infinity as $r$ ($p$?) tends to zero. At this stage you can just substitute in $r$/$p$. (Although the expresssion will be different once you consider the above)
Otherwise I believe this is fine.
