Limit of a function with two variables Could anyone help me with step-by-step solution of this limit?
$$\lim_{x \to 2, y \to 3} \frac{3x-2y}{y-x-1} $$
I used a calculator that gave me $-3$ as answer.
Thanks a lot.
 A: The limit does not exist. First fix $y$, and let $x \to 2$ to get: $L = \lim_{y\to 3} \dfrac{6-2y}{y-3} = -2$. Then fix $x$, and let $y \to 3$ to get: $L = \lim_{x \to 2} \dfrac{3x-6}{2-x} = -3$. Thus $L$ has two different values ,which shows the limit does not exist.
A: Let $x=2+\delta$ and $y=3+\epsilon$. Then your quotient becomes
$$
\frac{3\delta-2\epsilon}{\epsilon-\delta}=\frac{3\frac\delta\epsilon-2}{1-\frac\delta\epsilon}
$$
Thus, the value of the quotient depends on the direction you approach $(2,3)$. Therefore, the limit does not exist.
A: Consider two paths: $y=2x-1$ and $y=3x-3$. Both of these satisfy the condition that when $x \to 2, y \to 3$. 
Path 1: $y=2x-1$.
$$\displaystyle \lim_{(x,y)\to (2,3)}\dfrac{3x-2y}{y-x-1} =\lim_{(x,y)\to (2,3)} \dfrac{3x-2(2x-1)}{(2x-1)-x-1} = -1$$
Path 2: $y = 3x-3$.
$$\displaystyle \lim_{(x,y)\to (2,3)}\dfrac{3x-2y}{y-x-1} =\lim_{(x,y)\to (2,3)} \dfrac{3x-2(3x-3)}{(3x-3)-x-1} = \infty$$
Since the limits are not the same, then the limit does not exist. 
A: $$\lim_{(x, y)\to (2, 3)} \frac{3x-2y}{y-x-1} $$
Using polar coordinates, we have
$$ r\to\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$$
$$\phi\to\arctan\left(\frac{3}{2}\right) $$
So now we have
$$\lim_{r\to \sqrt{13}^+} \frac{3r\cos\phi-2r\sin\phi}{r\sin\phi -r\cos\phi-1} $$
$$=\lim_{r\to \sqrt{13}^+} \frac{3\cos\phi-2\sin\phi}{\sin\phi -\cos\phi-\frac{1}{r}} $$
This limit is clearly dependent on $\phi$, therefore this limit does not exist. 
