# Please halp :( (1 Viewer)

#### hayabusaboston

##### Well-Known Member
Find the limit as x and y approach 0 of

(7x^2y^2) / ( x^2+2y^4)

Using sandwich theorem

I haven't done this shit in ages and just can't brain how to do it right now

I feel like something to do with proving that fraction is less than 1 and clearly above 0 so the limit must be zero, but idk how to show it

Last edited:

Anyone?

#### lachlanwood

##### New Member
The fraction is greater than or equal to zero.
By the AM-GM inequality, the denominator is greater than or equal to

2root(2x^2y^4) = 2root(2)xy^2

Therefore, the fraction is less than or equal to

(7x^2y^2) / (2root(2)xy^2) = 7x/2root(2)

Lim x,y->0 (7x/2root(2)) = 0

Hence the limit of the fraction is greater than or equal to zero, but also less than or equal to zero, so it must be 0 by the squeeze theorem

#### InteGrand

##### Well-Known Member
The fraction is greater than or equal to zero.
By the AM-GM inequality, the denominator is greater than or equal to

2root(2x^2y^4) = 2root(2)xy^2

Therefore, the fraction is less than or equal to

(7x^2y^2) / (2root(2)xy^2) = 7x/2root(2)

Lim x,y->0 (7x/2root(2)) = 0

Hence the limit of the fraction is greater than or equal to zero, but also less than or equal to zero, so it must be 0 by the squeeze theorem
$\bg_white \noindent By the way, an easier way is to just say the denominator is greater than or equal to x^2, since 2y^4 \geq 0. Then carry on similarly.$

#### lachlanwood

##### New Member
$\bg_white \noindent By the way, an easier way is to just say the denominator is greater than or equal to x^2, since 2y^4 \geq 0. Then carry on similarly.$
Thanks!