Question

Need some help with limits, why the limit of sen(f(x))/f(x) when (x,y)->(0,0) is 1???

Answer 1

in English, do you mean
why is\[\lim_{x\to0}\frac{\sin x}x\]?

Answer 2

oh you have two variables, I see...

Answer 3

yes, I have two variables I read that is always = 1 but I really don't get why :S

Answer 4

it makes sense intuitively, right?
since we know the above is true
I'm just not sure how to prove that it's 1 in multivariable situations...
is it not sufficient to say that\[\lim_{(x,y)\to(0,0)}\frac{\sin f(x)}{f(x)}\]and since \(y=f(x)\) we have\[\lim_{y\to0}\frac{\sin y}y\]on which we can use lhospital's rule?
(probably not rigorous enough, sorry...)

Answer 5

U_U what you write is true but I need to know how is this true in multivariable situations :S

Answer 6

are you sure it aint? \[\lim _{(x,y)→(0,0)}sinf(x,y) \div f(x,y)\]

Answer 7

yes you wrote it correctly sorry :S

Answer 8

can you help me?

Answer 9

well, lets think about it, are we allowed to evaluate the limits one after another?

Answer 10

well if we have two variables yes, right?

Answer 11

ok, just read the definition of limit, it must "evenly" converge along any of the possibilities
so we could use lhospital in any of the cases

Answer 12

so I use L'hopital twice?

Answer 13

\[\lim _{(x,y)→(0,0)}sinf(x,y) \div f(x,y)\] , saly f(x,y)= x+y+1, it does not hold... since sen(1)/1 =not 1

Answer 14

is the f(x,y) given?

Answer 15

check the third post here:
http://www.physicsforums.com/archive/index.php/t-318232.html

Answer 16

ok I'll write the hole thing \[\lim_{(x,y) \rightarrow (0,0)}\sin (xy) \div xy\]

Answer 17

ahhh ok, in that case, just change to polar coordinates, the lim (x,y)->0,0 turns into lim(r)->0

Answer 18

Now I get it what you meant before about to use L'hopital in both cases I read that you can only use L'hopital in one variable but if I make that change I can use it thnx

Answer 19

what I meant is that you evaluate lim x->0 (f(x,y_0)) as if y_0 where constant, and for every y_0 notequal 0 the lim must converge, and then the same but with y,x_0

Answer 20

then again going to polar coordinates is the right thing to do here.

Answer 21

This is what I did because x,y -> 0,0 I just make x=y so I'll just have one variable and then I just use L'hopital, Am I wrong?

Answer 22

yes, it is wrong because you just tested one path, a function of 2 variables represents a surface, the lim must converge from every path to the point that its being tested

Answer 23

when transforming to polar coordinates, you will get a function of r,theta,gamma, but since the limit simplifies to only r, you can use l'hospital o solve it

Answer 24

mmm I see...

Answer 25

as a completely different argument it is also valid to state: f(x)= x and so is f(y)=y, are continuous functions around 0.
therefore f(x,y) =x*y is also continuous around 0,0, this implies that if one of the paths converges, all the others must also converge.

Answer 26

thnx really now I get it