# 2 multivariable limit questions.

## Homework Statement

Hello, I need help on the following questions, mainly to make sure i did them right.

for both, lim (x,y) -> (0,0)

1) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP3619hd2e84h398acc80000110c0eac25d415ha?MSPStoreType=image/gif&s=57&w=72&h=44 [Broken]

2) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP43819hd2e19gfb031c4000051d54bgf834ga83b?MSPStoreType=image/gif&s=47&w=80&h=40 [Broken]

## The Attempt at a Solution

1) so for this one I multiplied by the conjugate of the denominator.
So what i end up with is as follows

x(x-y)(sqrt x + sqrt y) / (x-y)

x-y cancels so I am left with

x(sqrt x + sqrt y)

so I get the limit as 0 when i plug in x and y.

2)
This one I am clueless about what to do. I keep getting an indeterminate form 0/0. I tried L'hopital's rule but it doesn't really help. Any ideas? Do i need to use some trigonometric identity here?

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member
(1) This is correct.

(2) Try approaching (0, 0) along some line which passes through the origin.

Alternatively, you can see what limit you get if you approach the x-axis along the line x = b, for some constant, b ≠ 0. Then, if that limit exists, approach (0,0) with that expression. Do similar for the y-axis.

SammyS
Staff Emeritus
Homework Helper
Gold Member
More on (1).

Of course you can only approach (0,0) from the first quadrant or the positive x or y axis.

I'm still a bit lost on #2. Using the first method such as taking a curve, say y=x that passes through the origin i still end up with 0/0, i tried L'hopitals rule after using y=x for example and I still don't get anywhere.

as for taking some x=b, so suppose b=1, then I would lim x,y -> 1,0. which also gives me an indeterminate form 0/0?

SammyS
Staff Emeritus
Homework Helper
Gold Member
I don't see the image of the problem any longer.

Sorry bout that. This is the question.

lim
x,y -> 0,0

[cos(xy)-1]/[(x^2)(y^2)]

SammyS
Staff Emeritus
Homework Helper
Gold Member
Sorry bout that. This is the question.

lim
x,y -> 0,0

[cos(xy)-1]/[(x^2)(y^2)]
Suppose y=mx.

$\displaystyle \frac{\cos(xy)-1}{x^2\,y^2}\to \frac{\cos(m\,x^2)-1}{m^2x^4}$

Now for L'Hôpital

Hmm looks like there is an error somewhere in the latex code, can't see what is written, but I'll try subbing in mx like you said and see what I can do from there with L'hopitals. I might have done something wrong then.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Apply L'Hôpital twice for these.

Using L'hopitals will make me get rid of the x from the denominator eventually but i think it will take more than 2 times. This is what i have so far.

using L'hopitals the first time I got:

-2mx sin(mx^2) /( 4x^3)(m^2)

which is still 0/0 as x -> 0

second time:

[-2m sin(mx^2) - 4m^2x^2 cos(mx^2)]/](12x^2)(m^2)]

which is still 0/0

unless i did something wrong, i can keep using l'hopitals to remove x from the denominator and should not give me 0/0 but it will be a total of 4 times I have to use it.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Using L'hopitals will make me get rid of the x from the denominator eventually but i think it will take more than 2 times. This is what i have so far.

using L'hopitals the first time I got:

-2mx sin(mx^2) /( 4x^3)(m^2)

which is still 0/0 as x -> 0
...
Cancel what you can before proceeding.

Cancel what you can before proceeding.
...right! stupid mistakes :(

thanks so much for the help!

SammyS
Staff Emeritus
Homework Helper
Gold Member
...
as for taking some x=b, so suppose b=1, then I would lim x,y -> 1,0. which also gives me an indeterminate form 0/0?
I get a similar result for this method.

SammyS
Staff Emeritus
Homework Helper
Gold Member