Limit of $\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$ 
Find limit of $ \underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$

How can I do that? It is interesting due to mathematica says that 
$$\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}} = 0 $$
but wolfram that limit doesn't exists. What is more I am not sure too about existance of limit due to
$$\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}} $$ doesn't exists too...
 A: Assuming $y\ne 0$,
\begin{align*}
&\lim_{(x,y)\to(0,0)}
\frac
{-\frac{x y}{2}+\sqrt{x y+1}-1}
{y \sqrt{x^2+y^2}}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}
\cdot
\frac
{\frac{x y}{2}+\sqrt{x y+1}+1}
{\frac{x y}{2}+\sqrt{x y+1}+1}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac
{\left(\sqrt{x y+1}\right)^2-\left(\frac{x y}{2}+1\right)^2}
{2y \sqrt{x^2+y^2}}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac
{(xy+1)-\left(\frac{x^2y^2}{4}+xy+1\right)}
{2y \sqrt{x^2+y^2}}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
-\frac{x^2y^2}{8y\sqrt{x^2+y^2}}\\[4pt] 
=\;&\lim_{(x,y)\to(0,0)}
-\frac{x^2y}{8\sqrt{x^2+y^2}}\\[4pt]
=\;&\;\;0\\[4pt]
\end{align*}
A: Using the Maclaurin formula
$$
\sqrt {A + 1} - 1 - \frac {A}2 \sim -\frac 18 A^2 [A \to 0], 
$$
we get
$$
\frac {\sqrt { xy+ 1} - 1 - \frac {xy}2}{ y \sqrt {x^2 + y^2}} \sim -\frac 18\cdot \frac {x^2 y^2} {y \sqrt {x^2 + y^2}}.
$$
Now let
$$
x = r \cos t, y =r \sin t, 
$$
then
$$
\frac  {x^2 y^2}{y \sqrt {x^2 + y^2}} = \frac {r^4 \sin^2 t \cos^2 t}{r^2 \sin t} = r^2 \sin t \cos^2 t \rightrightarrows 0 \, [r \to 0^+],
$$
hence the limit exists and equals $0$ [if the "equivalence" holds]. 
