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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.

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5 Answers 5

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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.

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

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

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  • $\begingroup$ 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 ? $\endgroup$
    – Angelo
    Jan 18, 2023 at 22:12
  • $\begingroup$ @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 $1.$ $\endgroup$ Jan 19, 2023 at 0:27
  • $\begingroup$ But $\;\sin(x^2)+\sin(y^2)\neq\sin^2\!x+\sin^2\!y$. $\endgroup$
    – Angelo
    Jan 19, 2023 at 6:56
  • $\begingroup$ @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. $\endgroup$ Jan 19, 2023 at 7:52
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$\cos x=1-(x^2/2)$ when $x$ is small and similarly for $\cos y$. Substituting you get that the limit equals $1/2$.

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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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