Question

Need help working through a limit problem (showing that the limit is zero)

Answer 1

Evaluate \[\lim_{(x,y)\to (0,0)}\frac{3x^5-xy^4}{x^4+y^4}\]

Answer 2

\[0\leq\frac{|3x^5-xy^4|}{x^4+y^4}\]It looks like I might be able to use the triangle inequality somehow here to find a greater function that splits the numerator and then I could use\[\frac{x^4}{x^4+y^4}\leq 1\]\[\frac{y^4}{x^4+y^4}\leq 1\]to greatly simplify my work. I'm having trouble identifying this though.

Answer 3

\[L=\lim x\times\left({3x^4\over x^4+y^4}-{y^4\over x^4+y^4}\right)\]

Answer 4

@electrokid Not sure how that helps.
That isn't much different than saying \[
\lim_{(x,y)\to (0,0)}\frac{3x^5-xy^4}{x^4+y^4} = \lim_{(x,y)\to (0,0)}(3x^5-xy^4)\frac{1}{x^4+y^4}
\]

Answer 5

@wio yes. I did not do much there. :) @richyw was looking to split in the way above and thats just what I did in the parathesis.

Answer 6

here is how you can approach..
using L'hopital's rules,
\[L=\lim_{(x,y)\to(0,0)}{f_x\over g_x}=\lim_{(x,y)\to(0,0)}{15x^4-y^4\over4x^3}\\
=\lim_{(x,y)\to(0,0)}\left({15\over4}x-{y^4\over4x^3}\right)=-{1\over4}\lim_{(x,y)\to(0,0)}{y^4\over x^3}=\color{red}{\Large0}\]

Answer 7

I don't think you can use l'hopital since this is not just a function of x

Answer 8

sure can

Answer 9

in what case?

Answer 10

but you have to check that you get the same limits for both the partial derivatives.

Answer 11

@wio aint I right?

Answer 12

what if it was like \[\lim_{(x,y)\to (0,0)}\frac{(xy)^4}{(x^2+y^2)^3}\]

Answer 13

lHopital works on multi-variable limits, but I'm not sure of the rules.

Answer 14

I'm pretty sure what you typed is very very wrong

Answer 15

for the latter problem:
\[L=\lim_{(x,y)\to(0,0)}{f_x\over g_x}=\lim_{(x,y)\to(0,0)}{4x^3y^4\over3(x^2+y^2)^2(2x)}\longrightarrow0 \;\text{(I extrapolate)}\]

Answer 16

nope see that's where it fails.

Answer 17

numerator approaces "0" faster than the denominator

Answer 18

so, what limit did you get for the second problem?

Answer 19

o'course, similarly proceeding, we'd have to show the same limit is approached if we differentiate with "y"

Answer 20

yeah, but then clearly l'Hoptials rule does not work for multivariable functions because this limit clearly does not exist

Answer 21

so for my original limit I think the method I was doing before is more on the right track

Answer 22

Wait are we showing the limit is 0 or that it doesn't exist?

Answer 23

ok, for the second one, convert it to polar co-ordinates
\[L=\lim_{(r,\theta)\to(0,0)}\frac{(r^4\cos^4\theta)(r^4\sin^4\theta)}{r^5}=0\]

Answer 24

so, the limit exists here :)

Answer 25

@electrokid How did you decide that \(\theta\) goes to 0?

Answer 26

it doesn't, that's the whole point of using polar coordinates.

Answer 27

ok, fine.. lets leave theta to be something... but you seee, the "r" goes to 0 which multiplies the entire thing.
0 times anthing (including infinity) is zero

Answer 28

the limit still stands good

Answer 29

you will end up with different limits depending on the direction you approach the origin and thus the limit does not exist.

Answer 30

do not know if this can be trusted for multivariables.. but
http://wolfr.am/Y9r0tP

Answer 31

@richyw which path are you using to get it not to exist?

Answer 32

oops, I made a typo on the function I was using as an example

Answer 33

@Mertsj @.Sam.

Answer 34

I meant for y^4 in the denominator

Answer 35

but l'hoptial doesn'r work still

Answer 36

ehm.. :)
then, the limits of partial differentials with "x" and "y" do not match :)
and hence the limit should not exist

Answer 37

the partials diff with x -> 0
the partial diff with y -> not 0

Answer 38

What is the actual function we're working with?

Answer 39

I did not solve.. but I presume..

Answer 40

the first one. I wanted to use squeeze theroem

Answer 41

I just really don't trust this L'hopital thing. I've just never seen it before and surely oone of my textbooks would have mentioned it.

Answer 42

I just can't find an example that shows it doesn't work...

Answer 43

hehe.
looks like I invented/discovered something here :P

Answer 44

ok what about this. \[\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2}\]

Answer 45

http://arxiv.org/pdf/1209.0363.pdf
sadly, somebody has already published it

Answer 46

rule of thumb: when you find "x^2+y^2" -> go to polar

Answer 47

I know how to do this. I don't actually care if l'hoptial works or not. I can worry about it after the exam. I want to go in with what I have already learned.

Answer 48

this one .. does not exist
if you take path "y=x" -> limit is "2"
if you take path "y=x^2" -> limit is "0"

they do not match

Answer 49

so, limit does not exist

Answer 50

finding a larger function that goes to 0 will work.