<s>[INST] Evaluate $ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} $ 
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x +
 \frac{y^2}{2} \right )}{x^2+y^2} $$

I've a few doubts about this limit. I mean, if I take polar coordinates, I get that the limit doesn't exist. And Wolfram agrees with me. Even though, I've found a solution of this problem that doesn't say the same thing; which I transcribe next:
· The Taylor's second order polynomial of $ e^{x+y^2} $ in $ (0,0) $ is $ 1 + x + \frac{1}{2}x^2 + y^2 $.
· The Taylor's second order polynomial of $ \sin \left ( x + \frac{y^2}{2} \right ) $ in $ (0,0) $ is $ x + \frac{1}{2}y^2 $.
Then:
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} = \lim_{(x,y)\to(0,0)} \frac {1 + x + \frac{1}{2}x^2 + y^2-1-(x + \frac{1}{2}y^2)}{x^2+y^2} = \frac {1}{2} $$
Where's the mistake in this?

What I've tried so far:
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} $$
Let be $ \rho \geq 0 $ and $ \varphi \in [0,2\pi) $, so:
$$ \left\{\begin{matrix}
x = \rho\cos(\varphi)\\ 
y = \rho\sin(\varphi)
\end{matrix}\right. $$
Then:
$$ \begin{align*}
\lim_{\rho\to 0} \frac {e^{\rho\cos(\varphi)+(\rho\sin(\varphi))^2}-1-\sin \left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{(\rho\cos(\varphi))^2+(\rho\sin(\varphi))^2} &= \lim_{\rho\to 0} \frac {\rho\cos(\varphi)+(\rho\sin(\varphi))^2-\left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{(\rho\cos(\varphi))^2+(\rho\sin(\varphi))^2} \\ 
 &= \lim_{\rho\to 0} \frac {\rho\cos(\varphi)+(\rho\sin(\varphi))^2-\left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{\rho^2} \\ 
 &= \lim_{\rho\to 0} \frac {(\rho\sin(\varphi))^2- \frac{(\rho\sin(\varphi))^2}{2}}{\rho^2} \\ 
 &= \lim_{\rho\to 0} \frac {1}{2}\sin^2(\varphi)
\end{align*} $$
Which limit doesn't exit, because the result depends of $ \varphi $ which varies in $ [0,2\pi) $.
And here you can see that Wolfram agrees.

According to a comment below, I've calculated a third order Taylor's polynomial (respecto to $ \rho$) of $ e^{\rho\cos(\varphi)+(\rho\sin(\varphi))^2} $. I found that any Taylor's polynomial (in $ \rho = 0 $) of order grater than 2 is exactly $ 1+\rho\cos(\varphi)+(\rho\sin(\varphi))^2 $.
 [/INST] For $(x,y)$ near $(0,0),$ we have
$$e^{x+y^2} = 1 + (x+y^2) +(x+y^2)^2/2 + O((x+y^2)^3), \sin (x+y^2/2) = (x+y^2/2) +O((x+y^2/2)^3).$$
So the numerator of our expression is
$$(y^2/2 + x^2/2) +  xy^2 + y^4/2 + O((x+y^2)^3) + O((x+y^2/2)^3)= (y^2/2 + x^2/2) + r(x,y).$$
Dividing by $x^2 + y^2$ we get $1/2 + r(x,y)/(x^2+y^2).$ Now $r(x,y)/(x^2+y^2)\to 0.$ I'll leave this to you for now. Thus the desired limit is $1/2.$
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