What is the flaw in both of my approaches to this limit? I have been solving a 3D limit problem: $$\lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$which apparently, is $2$.  However, I cannot find the flaw in my approach.  Here it goes:
$$\mathrm{Let \space y=mx} \\\lim_{(x,y) \to (0,0)} \frac{x^2+m^2x^2}{\sqrt{x^2+m^2x^2+1}-1}= \lim_{(x,y) \to (0,0)} \frac{x^2(1+m^2)}{\sqrt{2x^2+1}-1}=\frac{0}{0} \\ \mathrm{Apply \space L'Hôpital's \space rule}\\ \lim_{(x,y) \to (0,0)}\frac{2x(1+m^2)}{(\frac{2x}{\sqrt{2x^2+1}})}$$
I've solved this two different ways and I keep getting the answer as $(1+m^2)$.  The Student Solutions Manual, alas, says the answer it $2$.  Just in case my mistake lies somewhere after this step, even though I got the same answer solving it two different ways after this point I'll give you one of my methods, the shorter of the two:
$$...\space=\lim_{(x,y) \to (0,0)} \frac{2x(1+m^2)\sqrt{2x^2+1}}{2x}=\lim_{(x,y) \to (0,0)}\frac{(1+m^2)\sqrt{2x^2+1}}{1}=(1+m^2)$$
No idea how to get to $2$ here.  Also, am I abusing notation by leaving the bounds of the limits in terms of $(x,y)$ rather than reducing it to $x$ when I have let $y$ be equal to some terms of $x$?

Rationalize the denominator:
\begin{equation}
\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}\cdot \frac{\sqrt{x^2+y^2+1}+1}{\sqrt{x^2+y^2+1}+1}
\end{equation}
\begin{equation}=\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{(x^2+y^2+\sqrt{x^2+y^2+1}-\sqrt{x^2+y^2+1}+1)-1}
\end{equation}
\begin{equation}=\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{x^2+y^2}
\end{equation}
\begin{equation}=\sqrt{x^2+y^2+1}+1 \, .
\end{equation}
