Evaluate Limit two variables , how to tackle these questions? I'm aware this is a beginner question :D but im new to the subject.
I thought about known limits like $\tan(x)/x$ as $x\to0$ but im stuck, I can't find a way to include both known limits because of the denominator... maybe using the fact that $|\sin x|<|x|<|\tan x|$ ? ...
a hint would be much appreciated :D
$$
\displaystyle
\lim\limits_{\substack{x\to 2 \\ y\to 1}}
\frac{\tan(y-1)\sin^2(2 y -  x)}{(x-2)^2 +(y-1)^2}=
$$
furtermore, any tips on how do i tackle these types of questions?
 A: Let consider for simplicity the change of variable $u=x-2 \to 0$ and $v=y-1 \to 0$ then
$$\lim_{(x,y)\to(2,1)}
\frac{\tan(y-1)\sin^2(2 y -  x)}{(x-2)^2 +(y-1)^2}=\lim_{(u,v)\to(0,0)}
\frac{\tan (v)\sin^2(2v-u)}{u^2 +v^2}$$
and then use
$$\frac{\tan (v)\sin^2(2v-u)}{u^2 +v^2}=\frac{\tan (v)}{v}\frac{\sin^2(2v-u)}{(2v-u)^2}\frac{v(2v-u)^2}{u^2 +v^2}$$
which can be solved using standard limits for the first two terms and polar coordinates for the last one.
A: There are three cases to consider when doing these limits.
Firstly, fix $x$ at its limit point, so $x=2$ and take $y\to 1$. We get:
$$\lim_{y=1}\frac{\tan(y-1)\sin^2(2y-2)}{(y-1)^2}=0$$
Secondly, fix $y$ at its limit point, so $y=1$ and take $x\to 2$. We get:
$$\lim_{x\to 2}\frac{\tan(0)\sin^2(2-x)}{(x-2)^2}=0$$
Thirdly, let $y(x)$ be any function defined such that $x\to 2 \implies y(x)\to 1$
Then, you take $$\lim_{x\to 2}\frac{\tan(y(x)-1)\sin^2(2y(x)-x)}{(x-2)^2+(y(x)-1)^2}$$
and evaluate this general limit. If all the limits are the same, then that limit is the limit of the multivariate function, if there is a single exception, it has no limit.
In this case, the denominator is screaming at you to take $x-2=r\cos\theta, y-1=r\sin\theta$, as $$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{r\to 0}f(r\cos\theta, r\sin\theta)$$
A: Just a note:
If you go by the @user's method, when you obtain the factor $$\lim_{(u,v)\to (0,0)}\frac{v(2v-u)^2}{u^2+v^2},$$ you can avoid polar coordinates, because
$$\begin{aligned}\left|\frac{v(2v-u)^2}{u^2+v^2}\right|&=\frac{|v|\cdot|2v-u|^2}{u^2+v^2}\\&\le\frac{|v|(2|v|+|u|)^2}{u^2+v^2}\\&\le\frac{|v|\cdot 4(|u|+|v|)^2}{u^2+v^2}\\&\le\frac{|v|\cdot 4((|u|+|v|)^2+(|u|-|v|)^2)}{u^2+v^2}\\&=\frac{|v|8(u^2+v^2)}{u^2+v^2}\\&=8|v|\end{aligned}$$
Now squeeze.

I used the triangle inequality:
$$|2v-u|=|2v+(-u)|\le 2|v|+|u|,$$then $$2|v|+|u|\le 2(|v|+|u|).$$
On the other hand: $$(|u|+|v|)^2\le(|u|+|v|)^2+\underbrace{(|u|-|v|)^2}_{\ge 0}=2(|u|^2+|y|^2)$$ or you can use Cauchy-Schwarz-Bunyakovski to prove the same inequality as here.
