Question about double limit

I am practising the resolution of double limits after quite of a long time of not doing many calculations. Most of the exercises I've tried didn't pose a problem, usually the squeeze theorem, the polar method or choosing carefully a pair of paths that lead to different limits do the trick.

However, now I am stuck with the following limit, and I want to not use derivatives (so no L'Hôpital)

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

Choosing the path $y = -x$ shows that, if the limit exists, it must be $+\infty$. I'd like to find a path which leads to a different limit. The path $y=x$ leads to $-\infty$, but I showed it using L'Hôpital and, as I said, I'd like to solve the limit wihout the use of derivatives. This is bugging me as I feel there is a convenient path that gives a different result, but I can't see it.

Can you help?

Solution 1:

With $x=y$ you want to compute $$ \lim_{x\to0}\left(\frac{1}{1-\cos x}+\frac{2}{\sin^2x}\right) $$ which definitely doesn't require l'Hôpital: the second fraction can be rewritten as $$ \frac{2}{1-\cos^2x} $$ and so you get $$ \lim_{x\to0}\frac{1+\cos x+2}{1-\cos^2x}=\lim_{x\to0}\frac{3+\cos x}{\sin^2x} $$ to which you cannot apply l'Hôpital.

With $x=-y$ you'd get $$ \lim_{x\to0}\left(\frac{1}{1-\cos x}-\frac{2}{\sin^2x}\right)=\lim_{x\to0}\frac{1+\cos x-2}{1-\cos^2x}=\lim_{x\to0}-\frac{1}{1+\cos x} $$ which is not $-\infty$