Limit of several variable function using Taylor expansion

Question

I need to find the following limit:

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

My approach to find the limit is to first plug in the point, this doesn't work as the function is not continuous at that point. My next step is to notice that there are functions of $\cos$ of $x$ and of $y$ separately which means that I can Taylor expand and this is a very close approximation to the original function and we can therefore treat it as the actual function. So I Taylor expand both $\cos x$ and $\cos y$ to two terms to get the new limit:

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

This simplifies to :

$$ \lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\frac{x^2+y^2}{2}-\frac{x^2y^2}{4} }$$

$$ \lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{x^2+y^2} \frac{1}{\frac{1}{2}-\frac{x^2y^2}{4(x^2+y^2)} }$$

the term

$$ -\frac{x^2y^2}{4(x^2+y^2)} $$ goes to zero and so the limit goes to $2$. Can someone check my work please, also I Taylor expanded both $\cos x$ and $\cos y$ to the same amount of terms, does it matter to how many terms I expand is there some unwritten rule about this since I notice that it affects the outcome. Also let me know if there is an alternative approach to solve this limit and how it is justified in order for me to get a better understanding of how to approach several variable limits. Thank you for your time and help :)

Answer 1

All your steps are correct, and the limit does converge to $2$. Although you only expanded to $2$ terms, the highest power of $x$ and $y$ in the denominator is $4$, compared to only $2$ in the numerator. So there is no problem with your work.

Side note: This is like simplifying a quotient of two rational expressions, where you can just take the highest coefficient of the numerator and denominator to find the limit at infinity. With the fraction $\frac{x^2y^2}{4(x^2+y^2)}$, $x^2$ and $y^2$ are multiplied on the top, but only added on the bottom, so the limit will tend to $0$.

Answer 2

Your answer is correct, but you should obtain it more rigourously, using Taylor-Young's formula. The computations are easier using polar coordinates, so set $x=r\cos\theta$, $y=r\sin\theta$. The denominator becomes \begin{align} 1-\cos(r\cos\theta) \cos (r\sin\theta)&=1-\bigl(1-\tfrac12r^2\cos^2\theta+o(r^2)\bigr)\bigl(1-\tfrac12r^2\sin^2\theta+o(r^2)\bigr) \\ &=\tfrac12 r^2(\cos^2\theta+\sin^2\theta)+o(r^2)\sim_{r\to 0}\tfrac12 r^2 \end{align} so that we have $$\frac{x^2+y^2}{1-\cos x\cos y}\sim_{r\to 0}\frac{r^2}{\frac 12r^2}=2.$$

Answer 3

Another Taylor-approach without polar coordinates could be as follows:

Having in mind that

• $\frac{1}{2}\left(\cos u + \cos v \right) = \cos \frac{u+v}{2}\cdot \cos \frac{u-v}{2}$

you may apply the coordinate transformation

• $(\star): x= \frac{u+v}{2}, y= \frac{u-v}{2}$

So, plugging this into the given expression you get \begin{eqnarray*}\frac{x^2+y^2}{1-\cos x\cos y} & \stackrel{(*)}{=} & \frac{u^2+v^2}{2-(\cos u + \cos v)} \\ & \stackrel{Taylor}{=} & \frac{u^2+v^2}{2-(1-\frac{u^2}{2} + o(u^2)+ 1-\frac{v^2}{2} + o(v^2))}\\ & = & \frac{1}{\frac{1}{2} + \frac{o(u^2)+o(v^2)}{u^2+v^2}} \\ & \stackrel{(\star\star): (u,v)\to (0,0)}{\longrightarrow} & \frac{1}{\frac{1}{2} + 0} = 2 \end{eqnarray*}

The last step $(\star\star)$ is justified as follows: $$ \left|\frac{o(u^2)+o(v^2)}{u^2+v^2} \right|\leq \left|\frac{o(u^2)}{u^2} \right| + \left|\frac{o(v^2)}{v^2} \right| \stackrel{(u,v)\to (0,0)}{\rightarrow} 0 $$