Wolfram Alpha says it doesn't have a limit, but I don't know how to demonstrate...

TurtleCat
TurtleCat

Wolfram Alpha says it doesn't have a limit, but I don't know how to demonstrate it

Please help.

All urls found in this thread:
http://math.stackexchange.com/questions/177067/lhospital-rule-for-two-variable
https://www.google.com.sa/search?q=z+%3D+%28x%5E3%2By%5E2%29%2F%28abs%28x%29%2Babs%28y%29%29&oq=z+%3D+%28x%5E3%2By%5E2%29%2F%28abs%28x%29%2Babs%28y%29%29&gs_l=serp.3...4099.8297.0.8625.15.15.0.0.0.0.304.1867.0j6j2j1.9.0.csbd%2Ccfro%3D1%2Csrna%3D10%2Csrs%3D20...0...1.1.64.serp..9.0.0.CkE4aWGPVQg
RumChicken
RumChicken

@TurtleCat
you shouldve paid attention during class you fucking faggit. go to office hours or make a friend.

StonedTime
StonedTime

@RumChicken
opiates check 'em

Spazyfool
Spazyfool

bump

Stark_Naked
Stark_Naked

Wait, shouldn't the limit be zero?

Raving_Cute
Raving_Cute

@Stark_Naked
Yep. There's a theorem that starts with an "L" OP. That's all the tips I'm giving you

massdebater
massdebater

@RumChicken
@StonedTime

Oh hi fag, you aren't going to deliver on your Western Digital Passports post, are you?

Methshot
Methshot

@TurtleCat

I had this exercise some years ago …
I don't remember the details, but I think there was something about the limit being different depending on the direction from which you tend toward zero.
Thus no limit in R².

(On second thought, it was probably another exercise.)

Flameblow
Flameblow

@Methshot
I've actually never done multi-variable limits, but the only thing that could change with direction in the equation would be x^3, which should have a smaller absolute value than y^2 as both approach zero, right?

WebTool
WebTool

@Flameblow
Yeah this function is too symmetric for what I was thinking about.

ZeroReborn
ZeroReborn

|x|+|y| >= |x+y|
|x3 + x2| <= |x2 + y2| near zero
thus lim(|stuff|) <= lim |x+y| = 0

More or less.
(I have no idea how the theorems are called in english, redact it yourself.)

Garbage Can Lid
Garbage Can Lid

While the numerator and denominator both go to zero, the numerator goes to zero much faster, so the limit is zero.

(1+1)/(1+1)
(0.001+0.01)/(0.1+0.1)
(0.000001+0.0001)/(0.01+0.01)
etc.

Playboyize
Playboyize

@Raving_Cute
L'op
lol

TechHater
TechHater

Print this out and turn it in.

Methshot
Methshot

Use l'hopitals rule by takeing the derivitive. Thus observing the rate of change, you will have to do so multiple times probubly. If terms dont cancel out. You can then observe which terms grow faster and then take the limit.

Pay attention to this part since it will help with understanding cool shit like algorithm analysis

Firespawn
Firespawn

TOO SMART >:^(

lostmypassword
lostmypassword

for 3space limits at a certain (x,y) there may be separate limits for different lines of approach. if this is true, then there is "no limit" at those xy coordinates overall. This is similar to when the left and right limits of a 2space function for some x are not the same, meaning that there is no limit at that x overall.

Ignoramus
Ignoramus

@TurtleCat
Nigger, that's indeterminate form.

Plug in 0 for x and y and show that you approach 0 divided 0.

You use L'hospital's rule on it, and see where that gets you, another user mentioned it already

Should end up with (3x² + 2y) divided by (1+1) I think that's right, unless I'm forgetting important rules about absolute values.

Then you can plug in 0 for x and y again, and you get 0 as a limit.

3(0)² + 2(0) = 0

Wolfram alpha should have seen this, but I may have also made a mistake.

Nojokur
Nojokur

@Ignoramus
slight mistake, last part should be

[3(0)² + 2(0)] ÷ 2 = 0 ÷ 2 = 0

Fried_Sushi
Fried_Sushi

@Garbage Can Lid
Nope here you are calculating the limit for (x^3+x^2)/(2|x|)

idontknow
idontknow

@Ignoramus
L'Hospital's rule on functions with multiple variables ?

Carnalpleasure
Carnalpleasure

@idontknow
Is not allowed.

Ignore @Ignoramus

I'm not qualified to discuss multivariable limits.

http://math.stackexchange.com/questions/177067/lhospital-rule-for-two-variable

AwesomeTucker
AwesomeTucker

@Ignoramus
I'hopital's rule is a fucking godsend

cum2soon
cum2soon

math is gay

TechHater
TechHater

@Methshot
it will help with understanding cool shit like algorithm analysis
It doesn't help with understanding, but it helps with solving.

cum2soon
cum2soon

Wolfram Alpha is an ass.
The limit does exist, it's just telling you that in THE COMPLEX SPACE it doesn't exist.
See 3D graph: https://www.google.com.sa/search?q=z+%3D+%28x%5E3%2By%5E2%29%2F%28abs%28x%29%2Babs%28y%29%29&oq=z+%3D+%28x%5E3%2By%5E2%29%2F%28abs%28x%29%2Babs%28y%29%29&gs_l=serp.3...4099.8297.0.8625.15.15.0.0.0.0.304.1867.0j6j2j1.9.0.csbd%2Ccfro%3D1%2Csrna%3D10%2Csrs%3D20...0...1.1.64.serp..9.0.0.CkE4aWGPVQg

(might wanna use Chrome if it doesn't work on Firefox)

The graph at (0,0) only has a removable discontinuity.

happy_sad
happy_sad

@idontknow
@Fried_Sushi
@Nojokur
@Ignoramus
@TurtleCat
this fucking shit was a waste of half of my fucking semester
every fucking thing = 0
WOW
so amazing math
so useful!
fuck off i payed money for those credits

FastChef
FastChef

sage 2

Snarelure
Snarelure

yu

w8t4u
w8t4u

@Methshot
You can't use l'hopitals rule in multivariable calculus m8

Spazyfool
Spazyfool

@TurtleCat
The ratio of the orders of those functions (3/1) is greater than 1, so the function diverges.

PackManBrainlure
PackManBrainlure

@TurtleCat
fuck off op.
Now you have a
a lower bound
for "how many
morons there
are in Leeky Forums?"

but I won't
say you the answer
unless you say something
that is proof that you
really need the answer

Illusionz
Illusionz

from 40 post none is useful.
I'll say that there are 10-20 posters
but I have to say the right answer
because I have to.
There are <1/10 of math/physics related
undergraduates here.
8ch.org is for kiddos.

Now there are your answer
x=r cos a
y=r sin a
lim (x,y)->(0,0)=lim r->0
(x3+y2)/|x|+|y|= (r3(cos a)3+r2(sin a)2)/(r(|cos a| + |sin a|))<r
so
limit is 0.

askme
askme

I will explain a little better
|cos a|+|sin a| <2
but by triangle inequality
1<|cos a| +|sin a|
if r > 1
r^2(cos a)^3+r (sin a)^2)/(|sin a| + |cos a| ) < r(r cos^3+sin^2) < r(r+1)

because
and r cos^3 < r
and sin ^2 <= 1

Inmate
Inmate

@askme
if r > 1
I made a mistake
if r< 1 but it doesn't matter.

WebTool
WebTool

@askme
Using polar coordinates is lazy m8

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