Use rationalization to simplify the given expression in part (a). Then find the indicated limit in part (b).

Question

(a.)
3y2/(the square root of (y2 + y + 1))−(the square root of (y + 1))

b.)lim y>0 3y2/(the square root of (y2 + y + 1))−(the square root of (y + 1))

Answer 1

Do you mean

[3y^2/√(y^2+y+1)] - √(y+1)

or

3y^2 / [√(y^2+y+1) - √(y+1)]

?

Answer 2

the first one

Answer 3

no I mean the second one
3y^2 / [�ã(y^2+y+1) - �ã(y+1)] sorry

Answer 4

I thought so, but thought I'd check.

1/[√(y^2+y+1) - √(y+1)]

multiply top and bottom by

√(y^2+y+1) + √(y+1) and you have

[√(y^2+y+1) + √(y+1)]/(y^2+y+1 - (y+1))
= [√(y^2+y+1) + √(y+1)]/(y^2-1)

So now, a couple of applications of l'Hospital's Rule will get rid of that pesky 0/0.