<s>[INST] $\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.
 [/INST] 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 ,
$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}=$
$=\dfrac12\!\cdot\!\!\lim\limits_{(x,y)\to(0,0)}\dfrac{\sin^2\left(\frac{x+y}2\right)+\sin^2\left(\frac{x-y}2\right)}{\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}}\underset{\overbrace{\text{by letting }u=\frac{x+y}2\text{ and }v=\frac{x-y}2}}{=}$
$=\dfrac12\!\cdot\!\!\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}=\dfrac12\!\cdot\!1=\dfrac12\;.$
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