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

Please help.

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TurtleCat

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

Please help.

17 months ago

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

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

@TurtleCat

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

17 months ago

Spazyfool

bump

17 months ago

Stark_Naked

Wait, shouldn't the limit be zero?

17 months ago

Raving_Cute

@Stark_Naked

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

17 months ago

massdebater

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

17 months ago

Methshot

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.)

17 months ago

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?

17 months ago

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.)

17 months ago

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.

17 months ago

TechHater

Print this out and turn it in.

17 months ago

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

17 months ago

Firespawn

TOO SMART >:^(

17 months ago

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.

17 months ago

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.

17 months ago

Fried_Sushi

@Garbage Can Lid

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

17 months ago

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

17 months ago

cum2soon

math is gay

17 months ago

TechHater

@Methshot

it will help with understanding cool shit like algorithm analysis

It doesn't help with understanding, but it helps with solving.

17 months ago

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.

17 months ago

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

16 months ago

FastChef

sage 2

16 months ago

Snarelure

yu

16 months ago

Spazyfool

@TurtleCat

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

16 months ago

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

16 months ago

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.

16 months ago

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

16 months ago