Does $\lim_{(x,y)\to (a,b)} \frac{f(x)}{g(x,y)}$ that equals to $\lim_{(x,y)\to (a,b)} \frac{g(x,y)}{f(x)}$ mean that the limit equals to 1? Using L'hopital rule: $\lim_{x \to x_i, y \to y_i} \frac{x-x_i}{\sqrt{(x-x_i)^2 + (y-y_i)^2}} = \lim_{x \to x_i, y \to y_i} \frac{\frac{d(x-x_i)}{dx}}{\frac{d\sqrt{(x-x_i)^2 + (y-y_i)^2}}{dx}} = \lim_{x \to x_i, y \to y_i} \frac{1}{\frac{x-x_i}{\sqrt{(x-x_i)^2 + (y-y_i)^2}}} = \lim_{x \to x_i, y \to y_i} \frac{\sqrt{(x-x_i)^2 + (y-y_i)^2}}{x-x_i} = 1?$
When a limit equals itself$^{-1}$, is that limit equal to 1?
 A: Hint: when we have $a = \frac{1}{a}$, then it gives $a^2 = 1$.
A: Firstly observe that by $x-x_i=u$ and $y-_i=v$ we have
$$\lim_{x \to x_i, y \to y_i} \frac{x-x_i}{\sqrt{(x-x_i)^2 + (y-y_i)^2}} =\lim_{(u,v)\to (0,0)} \frac{u}{\sqrt{u^2 + v^2}} $$
which doesn't exist (just consider path $u=v=t\to 0^+$ and path $u=v=t\to 0^-$).
For the general question when
$$\lim_{(x,y)\to (a,b)} F(x,y)=L\neq 0$$
exists then by reciprocal law for limits
$$\lim_{(x,y)\to (a,b)} \frac{1}{F(x,y)}=\frac 1L\neq 0$$
and thus when both limits exist
$$\lim_{(x,y)\to (a,b)} F(x,y)=\lim_{(x,y)\to (a,b)} \frac{1}{F(x,y)}$$
implies $L^2=1$.
A: L'hopital rule is for one variable not multiple variables,
you have a problem lim (x->x0) lim(y->y0) f(x,y) =/= lim(y->y0) lim (x->x0) f(x,y)
in the first case the limit is 1 and in the second case lim(y->y0) sqrt((y-y0)^2)/0 which is infinite.
So the answer is no, the limit for multiple variables you need to transform those variables into the polar coordinates and proof that a limit doesn't depend on particular a vector direction.
