# Homework Help: Sandwich Theorem proof lim (x,y) -> (0,0) (4x^3*y^4)/(3y^4+5x^8)

1. Mar 23, 2012

### Phyrrus

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

lim (x,y) -> (0,0) (4x^3*y^4)/(3y^4+5x^8)

When y=kx -> lim x->0 = 0
The limit definitely is zero, but I can't prove it.

3. The attempt at a solution

? < lim (x,y) -> (0,0) (4x^3*y^4)/(3y^4+5x^8) < lim (x,y) -> (0,0) (4|x^3|*y^4+(15/4)|x^3|x^8)/(3y^4+5x^8) = (3/4)|x^3|

How do you find the lower limit of such a proof?
Can it be said that lim (x,y) -> (0,0) (4x^3*y^4)/(3y^4+5x^8) is equal to lim (x,y) -> (0,0) (|4x^3|*y^4)/(3y^4+5x^8) and therefore find the 'sandwich' of the new function?

2. Mar 23, 2012

### LCKurtz

Hint: Your denominator is non-negative. If you drop the $5x^8$ from the denominator, the fraction will be larger in absolute value than it is now.

3. Mar 23, 2012

### Phyrrus

Thanks, that is quite a simple solution. However, what can I use for the smaller valued function? Seeing as there is an x^3 in the numerator, I can't use 0 for the lower limit.

4. Mar 23, 2012

### Phyrrus

Oh wait, I just need to put in a negative sign for the lower limit... it's late

Cheers mate

5. Mar 23, 2012

### LCKurtz

You can if you put absolute value signs around the whole fraction. Remember if the absolute value of something goes to zero, then the something does.

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