2

Evaluation of $$\lim_{(u,v)\rightarrow (0,0)}\frac{v^2\sin(u)}{u^2+v^2}$$

We will calculate the limit along different path.

*Along $u$ axis, put $v=0$, we get limit $=0$

*Along $v$ axis, put $u=0$, we get limit$=0$

*Along $v=mu$ lime, we get $\displaystyle \lim_{u\rightarrow 0}\frac{m^2\sin(u)}{(1+m^2)}=0$

So the limit $$\lim_{(u,v)\rightarrow (0,0)}\frac{v^2\sin(u)}{u^2+v^2}=0$$

But walframalpha shows limit does not exists.

please help me where i am wrong . Thanks

jacky
  • 5,194
  • There are 2 var limits at the origin which go to zero along every line through the origin, but which don't exist because along some curved path is not zero. – coffeemath Aug 13 '20 at 07:02
  • Thanks coffeemath. Can you plesse explain me. Thanks – jacky Aug 13 '20 at 07:04

3 Answers3

6

The limit is $0$ indeed. If $(u,v)\ne(0,0)$, then$$0\leqslant\left|\frac{v^2\sin(u)}{u^2+v^2}\right|\leqslant|\sin(u)|\leqslant|u|$$and so, since $\lim_{(u,v)\to(0,0)}|u|=0$, it follows from the squeeze theorem that your limit is $0$ indeed.

However, your justification is wrong. It is not enough to show that the limit is $0$ if $(u,v)$ approaches $(0,0)$ along some paths to deduce that.

José Carlos Santos
  • 427,504
2

In order for a limit to exist, it has to be path independent, meaning no matter how the point (u,v) approaches (0,0), the limit exists and is always the same. You have just checked the $u$- and $v$-axis and every straight line into the origin. However, you could for example approach the origin in a spiral and might get a different result.

Mandelbrot
  • 555
  • Thanks Mandelbrot .Can you please explain me in detail. Thanks – jacky Aug 13 '20 at 07:04
  • Means the path along with limit non zero. – jacky Aug 13 '20 at 07:08
  • 1
    $\lim_{(u,v)\to (0,0)} f(u,v) = L\in \mathbb{R}$ is the same as saying that for every $\epsilon>0$ there exists an $\delta>0$ such that $\mid f(u,v)-L\mid <\epsilon$ for all $(u,v)$ in our domain that satisfy $\mid (u,v)-(0,0)\mid=\mid (u,v)\mid$. In other words, if your point $(u,v)$ approaches the origin, $f(u,v)$ should approach $L$.

    This is an equivalent way of saying that for every sequence $a_n$ with $a_n\to (0,0),n\to \infty,$ it is true that $f(a_n)\to L,n\to \infty$. So you have taken 2 sequences along the axes and all straight lines, which isn't enough.

     – Mandelbrot Aug 13 '20 at 07:09
1

As mentioned before, the limit should be zero for every path. But you can rewrite your function as following: $$ \frac{v^2\sin(u)}{u^2+v^2}=\frac{\sin(u)}{v^2(u^2+v^2)}=\frac{\sin(u)}{\frac{1}{v^2}(u^2+v^2)}=\frac{\sin(u)}{\frac{u^2}{v^2}+1} $$

Since we are calculating the limit $+1$ in the denominator really doesn't bother us. But we can see, that the function $(u/v)^2$ depends only on path, no mather how close to zero $u$ and $v$ are. So we can choose path, so that expression $\frac{\sin(u)}{\frac{u^2}{v^2}+1}$ has some non-zero constant value: $$ \frac{\sin(u)}{\frac{u^2}{v^2}+1}=A $$ $A$ can't be arbitrary big (at least I can say,that $A<1$), but than we can have path parameterised as: $$ v^2=\frac{u^2}{\sin(u)-A} $$ This path would than lead to non-zero limit.

Vid
  • 377
  • Ok, I youst found out, that I have problem, when $sin(u)<A$, which leads to imaginary $v$, any idea how to avoid this? – Vid Aug 13 '20 at 07:29