How prove this limit via epsilon delta? $$\lim_{(x,y) \to (4,1)}{\frac{y}{2x-y}}=\frac{1}{7}$$
I know so far that $|x-4|<\delta\quad \&\quad |y-1|<\delta$, and that I can use
$$\bigg|\frac{y-1}{2x-y} + \frac{1}{2x-y} - \frac{1}{7}\bigg|$$ and I get one delta, but how to continue from here, or is it even correct?
 A: To simplify we can change coordinates $u=x-4$ and $v=y-1$ such that
$$\lim_{(x,y) \to (4,1)}{\frac{y}{2x-y}}=\lim_{(u,v) \to (0,0)}{\frac{v+1}{2u-v+7}}=\frac{1}{7}$$
and assuming wlog $|u|,|v|\le 1$
$$\left|\frac{y-1}{2x-y}  - \frac{1}{7}\right|=\left|\frac{v+1}{2u-v+7}  - \frac{1}{7}\right|=\left|\frac{8v-2u}{7(2u-v+7)}\right|\le$$
$$\le\frac87\left|\frac{v}{2u-v+7}\right|+\frac27\left|\frac{u}{2u-v+7}\right|\le$$
$$\le \frac87\left|\frac{v}{4}\right|+\frac27\left|\frac{u}{4}\right| \le \frac87\frac{\delta}{4}+\frac27\frac{\delta}{4}=\frac{5}{14} \delta$$
A: I invariably give the same advice for these situations, it is easier to work in $(0,0)$ because it triggers more reflexes.
So set $\begin{cases}x=4+u & u\to 0\\y=1+v & v\to 0\end{cases}$
Note that you can as well have $|u|<\delta$ and $|v|<\delta$.
Then evaluate $f(x,y)-\frac 17=\dfrac{Num(u,v)}{Den(u,v)}$
Notice that $N(u,v)\to 0$ and $Den(u,v)\to cst\neq 0$
What you want to do is proving $|Num(u,v)|<n\delta$ for some $n$, use triangular inequality and $\delta<\epsilon$
And for denominator $|Den(u,v)>k|$ for that use $\delta<1$ then $|7+2u-v|>7-2-1=4$
Conclude by taking $\delta=\min(1,\epsilon)$ so that both parts are verified.
