Evaluating $\lim_{(x,y) \to (\pi,0)} {\cos(x) + 1 + {y^2/2} \over (x - \pi)^2 + y^2}$ I'm struggling to evaluate this limit problem:
$$\lim\limits_{(x,y) \to (\pi,0)} {\cos x + 1 + {y^2/2} \over (x - \pi)^2 + y^2}$$
I've tried this with paths $x=\pi$, $y=0$, $y=x-\pi$, ... and so far all of them have resulted in $1 \over 2$. However when I try to evaluate this in WolframAlpha, it says that the limit does not exist.
I haven't found any paths that result in a value other than $1/2$. How can I prove that this limit does not exist?
 A: Let $u=x-\pi$, then
$$\lim_{(x,y)\to(\pi,0)}\frac{\cos(x)+1+y^2/2}{(x-\pi)^2+y^2}
= 
\lim_{(u,y)\to(0,0)}\frac{-\cos(u)+1+y^2/2}{u^2+y^2}$$
Note that for small $u$, $\cos(u)=1-u^2/2+O(u^4)$, hence
\begin{align}
\lim_{(u,y)\to(0,0)}\frac{-\cos(u)+1+y^2/2}{u^2+y^2}
&=\lim_{(u,y)\to(0,0)}\frac{u^2/2+O(u^4)+y^2/2}{u^2+y^2}\\
&= \lim_{(u,y)\to(0,0)}\left[\frac12 + \frac{O(u^4)}{u^2+y^2}\right]
=\tfrac12
\end{align}
A: You should try $x' = x - \pi$ and then $$\lim\limits_{(x,y) \to (\pi,0)} {\cos x + 1 + {y^2/2} \over (x - \pi)^2 + y^2} = \lim\limits_{(x',y) \to (0,0)} {1-\cos x' + {y^2/2} \over (x')^2 + y^2}$$
$$=  \lim\limits_{(x,y) \to (0,0)} {1-\cos x \over x^2 + y^2} + \frac{1}{2} {{y^2} \over x^2 + y^2}$$
$$=  \lim\limits_{(x,y) \to (0,0)} { 2\sin^2(x/2) \over x^2 + y^2} + \frac{1}{2} { y^2 \over x^2 + y^2}$$
$$=  \lim\limits_{(x,y) \to (0,0)} \frac{\frac{1}{2}x^2}{x^2+y^2} \left ({\sin^2(x/2)\over (x/2)^2} \right) + {\frac{1}{2}y^2\over x^2 + y^2}$$
$$=  \lim\limits_{(x,y) \to (0,0)} \frac{\frac{1}{2}x^2}{x^2+y^2} \left (1+{\sin^2(x/2)\over (x/2)^2} - 1 \right) + {\frac{1}{2}y^2\over x^2 + y^2}$$
$$=  \lim\limits_{(x,y) \to (0,0)} \frac{1}{2}\frac{x^2 + y^2}{x^2+y^2} + \lim\limits_{(x,y) \to (0,0)} \frac{1}{2}\frac{x^2}{x^2+y^2} \left ( {\sin^2(x/2)\over (x/2)^2} - 1\right )  $$
$$= \frac{1}{2}  $$
A: By the change of variable


*

*$u=x-\pi\to 0$

*$v=y\to 0$
we have that
$$\lim\limits_{(x,y) \to (\pi,0)} {\cos(x) + 1 + {y^2/2} \over (x - \pi)^2 + y^2}=\lim\limits_{(u,v) \to (0,0)} {{1-\cos u+v^2/2} \over u^2+v^2}=\lim\limits_{(u,v) \to (0,0)} {{u^2/2+v^2/2+o(u^2+v^2)} \over u^2+v^2}=\lim\limits_{(u,v) \to (0,0)} \frac12 +\frac{o(u^2+v^2)}{u^2+y^2}
=\frac12$$
