How to evaluate the following multivariable limit?

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

lim (x,y)->(0,0) (cos(x) - 1 + (x^2/2)) / ( x^4 + y^4)

3. The attempt at a solution

Am I correct in assuming that I have to solve by looking for a path where the limit doesn't exist/is different to another path (given that the epsilon thing is beyond the scope of the course)?

I've tried finding it along the path y=0, x=0 and y=kx, and every time I've found it approaches infinity. Not sure if I've done anything wrong or not, but looking at a graph of the function, it looks like approaches along the y=0 and x=0 lines should give me different limits (according to the answer, the limit does not exist at 0,0).

Any help?

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

OK, I'm fairly sure I've gotten the above question right, but am stuck on another.

Find the limit of the following function using the Sandwich Theorem:
f(x,y) = (7x^2y^2)/(x^2 + 2y^4) as (x,y)->(0,0)

How would we go about it? Seeing as all terms are non-negative, zero will always be =<, but how about finding an function that's >= and also simpler to solve?

You couldn't multiple/divide it by a function of x and/or y, because it'd make it larger or smaller depending on if x and/or y were smaller or greater than zero, correct? And adding a function of x and/or y wouldn't make it any easier to solve...

Would it be possible to just add a negative scalar (e.g. -1) to the denominator, then substitute (x,y)=(0,0) to evaluate the limit? That way, you'd have a non-zero denominator and as both numerator and denominator are continuous, it'd be very simple to solve.

0 =< f(x,y) =< (7x^2y^2)/(x^2 + 2y^4 - 1)
lim 0 =< lim f(x,y) =< lim (7x^2y^2)/(x^2 + 2y^4 - 1)
Therefore, lim f(x,y) = 0.

Seems too simple, though. Am I missing something or is it a legitimate approach?
 
Last edited:
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Hmmm...thinking further on the question, would the substitution y=kx have a limit of zero, not infinity?

Making the substitution, I get:
(cos(kx) - 1 + (x^2/2)) / ( x^4 + (kx)^4)

Are we allowed to use L'Hopital's? If so:
(k^4*cos(kx))/(24 + 24k^4). Sub x=0 gives:
k^4/(24(1+k^4), which is a varying limit depending on what value of k you use.

Does the above make sense?
 
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OK, I'm fairly sure I've gotten the above question right, but am stuck on another.

Find the limit of the following function using the Sandwich Theorem:
f(x,y) = (7x^2y^2)/(x^2 + 2y^4) as (x,y)->(0,0)

How would we go about it? Seeing as all terms are non-negative, zero will always be =<, but how about finding an function that's >= and also simpler to solve?

You couldn't multiple/divide it by a function of x and/or y, because it'd make it larger or smaller depending on if x and/or y were smaller or greater than zero, correct? And adding a function of x and/or y wouldn't make it any easier to solve...

Would it be possible to just add a negative scalar (e.g. -1) to the denominator, then substitute (x,y)=(0,0) to evaluate the limit? That way, you'd have a non-zero denominator and as both numerator and denominator are continuous, it'd be very simple to solve.

0 =< f(x,y) =< (7x^2y^2)/(x^2 + 2y^4 - 1)
lim 0 =< lim f(x,y) =< lim (7x^2y^2)/(x^2 + 2y^4 - 1)
Therefore, lim f(x,y) = 0.

Seems too simple, though. Am I missing something or is it a legitimate approach?
 
what happens if x=0. i.e. if we approach the point (0,0) along the y-axis? then the lim is 0, right? why?

what happens if we approach it along x axis, i.e. when y=0.

[tex] f(x,0)=\frac{cosx-1-\frac{1}{2}x^2}{x^4}[/tex]

now what is this limit as x-->0 ?
 
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You're right - it is zero. I accidentally had cos(0) as 0 instead of 1. :redface:

For f(x,0), can't we use L'Hopital's to simplify it to cos(x)/24 = 1/24? Therefore the limits along y=0 and x=0 are different, so the limit doesn't exist at the origin?
 
You're right - it is zero. I accidentally had cos(0) as 0 instead of 1. :redface:

For f(x,0), can't we use L'Hopital's to simplify it to cos(x)/24 = 1/24? Therefore the limits along y=0 and x=0 are different, so the limit doesn't exist at the origin?
that's right!
 
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Thanks for your help, sm.
 

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