# $\lim_{(x,y)\to (0,0)}\frac{1-(\cos x)(\cos y)}{x^2+y^2}$

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.

For all that kind of problem use asymptotic development i.e. for $$\;\cos\;$$ around 0 here :

$$\cos(x)=1-x^2/2 +o_0(x^3)$$

$$\dfrac{1-(1-x^2/2+o(x^3))(1-y^2/2+o(y^2))}{x^2+y^2}=\dfrac{x^2/2+y^2/2 - (x^2y^2/4) +o(x^3y^2+y^2x^3)}{x^2+y^2}$$

$$\dfrac{1-(1-x^2/2+o(x^3))(1-y^2/2+o(y^2))}{x^2+y^2}=1/2+\dfrac{ - (x^2y^2/4) +o(x^3y^2+y^2x^3)}{x^2+y^2}$$

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.

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: $$\lim_{r\to 0}\frac{r^2-r^4\sin^2\theta\cos^2\theta}{2r^2}=\frac12$$

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 $$\lim_{(x,y)\to (0,0)}{1-\cos x\cos y\over x^2+y^2}={1\over 2}$$

• How would you calculate the limit $\lim\limits_{(x,y)\to(0,0)}\dfrac{\sin^2\!x+\sin^2\!y}{x^2+y^2}\;?$ Is there a shorter way than the one I have written in my answer ? Jan 18, 2023 at 22:12
• @Angelo We have $$\sin(x^2)+\sin(y^2)=2\sin[(x^2+y^2)/2]\cos [(x^2-y^2)/2]$$ On dividing by $x^2+y^2,$ the limit is $1.$ Jan 19, 2023 at 0:27
• But $\;\sin(x^2)+\sin(y^2)\neq\sin^2\!x+\sin^2\!y$. Jan 19, 2023 at 6:56
• @Angelo Sorry, I have to change my reading glasses. For$f(x,y)={\sin^2x\over x^2},$ and $g(x,y)=f(y,x),$ apply $(*)$ from my answer. Jan 19, 2023 at 7:52

$$\cos x=1-(x^2/2)$$ when $$x$$ is small and similarly for $$\cos y$$. Substituting you get that the limit equals $$1/2$$.

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 ,

$$\lim\limits_{(x,y)\to(0,0)}\dfrac{1-\cos x\cos y}{x^2+y^2}=$$

$$=\dfrac12\!\cdot\!\!\lim\limits_{(x,y)\to(0,0)}\dfrac{\sin^2\left(\frac{x+y}2\right)+\sin^2\left(\frac{x-y}2\right)}{\left(\frac{x+y}2\right)^{\!2}+\left(\frac{x-y}2\right)^{\!2}}\underset{\overbrace{\text{by letting }u=\frac{x+y}2\text{ and }v=\frac{x-y}2}}{=}$$

$$=\dfrac12\!\cdot\!\!\lim\limits_{(u,v)\to(0,0)}\dfrac{\sin^2\!u+\sin^2\!v}{u^2+v^2}=\dfrac12\!\cdot\!1=\dfrac12\;.$$