$\lim_{(x,y)\to (0,0)}\frac{1-(\cos x)(\cos y)}{x^2+y^2} $ I need to find the limit for $$\lim_{(x,y)\to (0,0)}\frac{1-(\cos x)(\cos y)}{x^2+y^2} $$
whether exist.
I use many example (ex:line, interated limit, half angle formula, ...), and I always get the answer $1/2$. However, this does not mean the limit is $1/2$. Thus, I want to use definition to show that limit in fact be $1/2$. Unfortunately, I tried many times, and using the inequality I know to show, I failed. Can someone give me a useful inequality to solve it ? Or, point out that I miss somewhere, and show that the limit does not exist.
Thanks.
 A: For all that kind of problem use asymptotic development i.e. for $\;\cos\;$ around 0 here :
$$ \cos(x)=1-x^2/2 +o_0(x^3)$$
So your expression become :
$$ \dfrac{1-(1-x^2/2+o(x^3))(1-y^2/2+o(y^2))}{x^2+y^2}=\dfrac{x^2/2+y^2/2  - (x^2y^2/4) +o(x^3y^2+y^2x^3)}{x^2+y^2}$$
$$ \dfrac{1-(1-x^2/2+o(x^3))(1-y^2/2+o(y^2))}{x^2+y^2}=1/2+\dfrac{ - (x^2y^2/4) +o(x^3y^2+y^2x^3)}{x^2+y^2}$$
Note  now that $$ |x^2y^2/4| \leq |(x^2+y^2)(x^2+y^2)/4) $$
I think from here you can end the proof.
A: The way I would approach this problem is to multiply and divide by $1+\cos x\cos y$, then use $\sin^2a+\cos^2a=1$. You will then get $$\lim_{(x,y)\to(0,0)}\frac{1-\cos^2x\cos^2y}{(1+\cos x\cos y)(x^2+y^2)}=\lim_{(x,y)\to(0,0)}\frac{1-1+\sin^2 x+\sin^2y-\sin^2x\sin^2y}{2(x^2+y^2)}$$
Then, assuming that the limits exist, $\sin^2x\approx x^2$. Using $x=r\cos\theta$ and $y=r\sin\theta$, your limit becomes:
$$\lim_{r\to 0}\frac{r^2-r^4\sin^2\theta\cos^2\theta}{2r^2}=\frac12$$
A: Assume that for functions $f(x,y)$ and $g(x,y)$ we have $$\lim_{(x,y)\to (0,0)}f(x,y)=\lim_{(x,y)\to (0,0)}g(x,y)=a$$
Then (a simple proof in the spoiler)
$$\lim_{(x,y)\to (0,0)}{f(x,y)\,x^2+g(x,y)\,y^2\over x^2+y^2}=a\quad (*)$$

 $$f(x,y)x^2+g(x,y)y^2=f(x,y)[x^2+y^2]+[g(x,y)-(x,y)]y^2$$ Hence $${f(x,y)x^2+g(x,y)y^2\over x^2+y^2} =f(x,y)+{[g(x,y)-f(x,y)]y^2\over x^2+y^2}$$ Therefore $$\left |{f(x,y)x^2+g(x,y)y^2\over x^2+y^2}-a\right |\le |f(x,y)-a|+|g(x,y)-f(x,y)|$$

Let $$f(x,y)={1-\cos x\over x^2}\cos y,\qquad g(x,y)={1-\cos y\over y^2}$$
Then $$\lim_{(x,y)\to  (0,0)}f(x,y)=\lim_{(x,y)\to (0,0)}g(x,y)={1\over 2}$$
Observe that $$1-\cos x\cos y=f(x,y)x^2+g(x,y)y^2$$
Hence applying $(*)$ gives
$$\lim_{(x,y)\to (0,0)}{1-\cos x\cos y\over x^2+y^2}={1\over 2}$$
A: $\cos x=1-(x^2/2)$ when $x$ is small and similarly for $\cos y$. Substituting you get that the limit equals $1/2$.
A: Another way to calculate the limit
$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}\;.$
First of all, we will calculate the following limit :
$\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}\;.$
Let $\;\varphi:\,]\!-\!\infty,+\infty[\to\Bbb R\;$ be the function defined as :
$\varphi(t)=\begin{cases}\dfrac{\sin^2\!t}{t^2}\quad&\text{ for any }\,t\in\,]\!-\!\infty,+\infty[\,\setminus\,\{0\}\\\\\;\;1&\text{ for }\,t=0\end{cases}$
It follows that $\;\lim\limits_{t\to0}\varphi(t)=\varphi(0)=1\,.$
Moreover, for all $\,(u,v)\in\Bbb R^2\setminus\{(0,0)\}\;$ it results that
$\begin{align}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}&=\dfrac{\big[\varphi(u)\!-\!\varphi(v)\big]\!\left(u^2\!-\!v^2\right)\!+\!\big[\varphi(u)\!+\!\varphi(v)\big]\!\left(u^2\!+\!v^2\right)}{2\left(u^2+v^2\right)}=\\
&=\dfrac{\varphi(u)-\varphi(v)}2\cdot\dfrac{u^2-v^2}{u^2+v^2}+\dfrac{\varphi(u)+\varphi(v)}2\;.\end{align}$
Since $\;\left|\dfrac{u^2-v^2}{u^2+v^2}\right|\leqslant1\;$ for all $\,(u,v)\in\Bbb R^2\setminus\{(0,0)\}\;$ and
$\lim\limits_{(u,v)\to(0,0)}\varphi(u)=\lim\limits_{(u,v)\to(0,0)}\varphi(v)=1\;,\;$ it follows that
$\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}=1\;.\quad\color{blue}{(*)}$
Now, we will calculate the limit
$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}\;.$
For all $\,(x,y)\in\Bbb R^2\setminus\{(0,0)\}\;$ it results that
$\begin{align}\dfrac{1-\cos x\cos y}{x^2+y^2}&=\dfrac{1-\cos(x+y)+1-\cos(x-y)}{2\left(x^2+y^2\right)}=\\
&=\dfrac{2\sin^2\left(\frac{x+y}2\right)+ 2\sin^2\left(\frac{x-y}2\right)}{4\left[\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}\right]}=\\
&=\dfrac12\!\cdot\!\dfrac{\sin^2\left(\frac{x+y}2\right)+\sin^2\left(\frac{x-y}2\right)}{\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}}\;.\end{align}$
Consequently ,
$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}=$
$=\dfrac12\!\cdot\!\!\lim\limits_{(x,y)\to(0,0)}\dfrac{\sin^2\left(\frac{x+y}2\right)+\sin^2\left(\frac{x-y}2\right)}{\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}}\underset{\overbrace{\text{by letting }u=\frac{x+y}2\text{ and }v=\frac{x-y}2}}{=}$
$=\dfrac12\!\cdot\!\!\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}=\dfrac12\!\cdot\!1=\dfrac12\;.$
