what is the limit of this as it approaches to inf?

Question

{ (root (4n^2+3n+3)) - 2n }

Answer 1

The answer is 3/4.

First, take the conjugate.
Then divide top and bottom by the highest power.

[(root (4n^2+3n+3)) - 2n] * ([(root (4n^2+3n+3)) + 2n]/[(root (4n^2+3n+3)) + 2n])

= lim n --> oo of 3n + 3/(sqrt(4n^2 + 3n + 3) + 2n

Now divide top and bottom by highest power of n. You'd divide the top by n and the bottom that's inside the squareroot by n^2 while 2n just gets divided by n because it's not under the squareroot.

From there you can conclude that the lim n--> oo is 3/4.

Answer 2

I assume n approaches infinity. If so then

sqrt(4n^2 +3n+3) -> sqrt(4n^2) = 2n thus

sqrt(4n^2+3n+3)-2n --> 0 as n apporaches infinity

Answer 3

zero, the 3n+3 becomes negligible over time, then root 4n^2 is just 2n then subtract 2n to get zero

Answer 4

what ever you want it to be. life is such a mystery. its amazing.