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2 multivariable limit questions.

  • #1

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?
 
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Answers and Replies

  • #2
SammyS
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(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
SammyS
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More on (1).

Of course you can only approach (0,0) from the first quadrant or the positive x or y axis.
 
  • #4
Thanks for the reply.

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
SammyS
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I don't see the image of the problem any longer.
 
  • #6
Sorry bout that. This is the question.

lim
x,y -> 0,0

[cos(xy)-1]/[(x^2)(y^2)]
 
  • #7
SammyS
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Sorry bout that. This is the question.

lim
x,y -> 0,0

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

[itex]\displaystyle \frac{\cos(xy)-1}{x^2\,y^2}\to \frac{\cos(m\,x^2)-1}{m^2x^4}[/itex]

Now for L'Hôpital
 
  • #8
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
SammyS
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Apply L'Hôpital twice for these.
 
  • #10
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
SammyS
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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.
 
  • #12
Cancel what you can before proceeding.
...right! stupid mistakes :(

thanks so much for the help!
 
  • #13
SammyS
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...
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.
 
  • #14
SammyS
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Now for the bad news.

This all looks great, but it doesn't prove the limit is -1/2 or whatever we got.

If any were different than -1/2, then the limit DNE.

This may be good enough for your course at this time.
 

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