# 2 multivariable limit questions.

1. Oct 10, 2011

### AndreTheGiant

1. The problem statement, all variables and given/known data

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]

3. 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: May 5, 2017
2. Oct 10, 2011

### SammyS

Staff Emeritus
(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.

3. Oct 10, 2011

### SammyS

Staff Emeritus
More on (1).

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

4. Oct 10, 2011

### AndreTheGiant

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?

5. Oct 10, 2011

### SammyS

Staff Emeritus
I don't see the image of the problem any longer.

6. Oct 10, 2011

### AndreTheGiant

Sorry bout that. This is the question.

lim
x,y -> 0,0

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

7. Oct 10, 2011

### SammyS

Staff Emeritus
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

8. Oct 10, 2011

### AndreTheGiant

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.

9. Oct 10, 2011

### SammyS

Staff Emeritus
Apply L'Hôpital twice for these.

10. Oct 10, 2011

### AndreTheGiant

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.

11. Oct 10, 2011

### SammyS

Staff Emeritus
Cancel what you can before proceeding.

12. Oct 10, 2011

### AndreTheGiant

...right! stupid mistakes :(

thanks so much for the help!

13. Oct 10, 2011

### SammyS

Staff Emeritus
I get a similar result for this method.

14. Oct 10, 2011

### SammyS

Staff Emeritus