limit in two dimensions of a function

Question

The problem is show that the limit $$\lim_{(x,y)\rightarrow (1,0)}\frac{x-1+{2y}}{y}=2$$

I have obtained the following but failed to complete the exercise.

$$\left | x-1 \right |\leq \sqrt{(x-1)^2+y^2}< \delta$$ and $$\left | \frac{(x-1+2y)}{y}-2 \right |=\left | \frac{x-1}{y} \right |=\frac{1}{y}\left | x-1 \right |$$

Thank you.

Answer 1

This is false. The limit does not exist. If you take limit along $x=1-\frac 1 n$ and $y=\frac 1 n$ you get the limit as $1$ whereas the limit along $x=1$ is $2$. [Note that $(1-\frac 1 n , \frac 1 n) \to (1,0)$ as $n \to \infty$. I first took limit along this sequence. Then I took limit along the sequence $(1,\frac 1 n)$ which also tends to $(1,0)$].

Answer 2

Note that

$$\frac{x-1+{2y}}{y}= \frac{x-1}{y}+2$$

and therefore the limit depends upon the ratio $\frac{x-1}{y}$ as $(x,y)\to (1,0)$.