<s>[INST] 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...
 [/INST] 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]. 
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