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In the following 2 examples I used 2 different tricks to prove the existence of the limit.

a) How can I check every path?

b) Are there other tricks that can come in handy to solve multivariable limits? Here were sandwich and substitution with polarcoordinates.

c) To prove that a limit exist using the definition how would it work? Is it reasonable? Could you make an example with the examples I gave

d) To prove that the limit doesn't exist I saw that you have to find two paths going through the critic point and calculate the limit. How can I make such guesses of the suitable paths that will contradict the existance of a limit. I mean, for hypotesis I try 3 paths, everyone gives me the same limit, so I think the limit exists and start to try to prove the existence, but actually I didn't consider a path that will show that there is not a limit.

1) Given the function $f(x,y)= \frac{(x-1)^2 \log(x)}{(x-1)^2 +y^2}$ that is defined on the set $\Omega=\{(x,y)\in\Re^2|x>0\}$ , prove if the limit $\lim_{(x,y)\to(1,0)}f(x,y)$ exists. And if it does, what is the value?

\begin{align} |f(x,y)|=\frac{|(x-1)^2|}{|(x-1)^2+y^2|} |\log(x)|\leq|\log(x)| \\\\ \end{align} Since $\lim_{x\to 1}\log(x)=0$ and $0\leq \lim_{(x,y)\to(1,0)}|f(x,y)|\leq \lim_{x\to 1}\log(x)=0$ then $\lim_{(x,y)\to(1,0)}f(x,y)=0$

The limit exists and is $0$. It's proven thanks to the sandwich theorem.

2) Given the function $f(x,y)=\frac{\sin(x^2+y^2)}{x^2+y^2}$ defined in $\Omega= \Re^2\backslash \{(0,0)\}$, prove if the limit $\lim_{(x,y)\to(0,0)}f(x,y)$ exists. And if it does, what is the value? \begin{align} x=r\cos(\phi), y=r\sin(\phi)\\\\ \lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}= \lim_{r\to 0}\frac{\sin(r^2)}{r^2}\\ \end{align} Since now we have only one variable, and we have $\frac{0}{0}$ we can solve it with hopital. \begin{align} \lim_{r\to 0}\frac{\sin(r^2)}{r^2}= \lim_{r\to 0}\frac{2r\cos(r^2)}{2r}=\lim_{r\to 0}\cos(r^2)=1 \end{align} The limit exists and is $1$. We managed to control every path with polarcoordinates.

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