How to evaluate the following multivariable limit?

[rafehi]

Homework Statement

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

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?

Homework Statement

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?

 

[rafehi]

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?

 

[rafehi]

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?

 

[sutupidmath]

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.

$$f(x,0)=\frac{cosx-1-\frac{1}{2}x^2}{x^4}$$

now what is this limit as x-->0 ?

 

[rafehi]

You're right - it is zero. I accidentally had cos(0) as 0 instead of 1.

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?

 

[sutupidmath]

You're right - it is zero. I accidentally had cos(0) as 0 instead of 1.

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!

 

[rafehi]

Thanks for your help, sm.