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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.