You are finding the limit as n → what?
Hello all,
I am trying to understand whats going on algebra wise here,
(6n^4+2n^3) / (7n-11^4)
from here i needed to eliminate the n^4 terms, so;
((6n^4+2n^3) / (7n-11^4)) * (1/n^4)/(1/n^4) ..
so far so good, but apparantly this all then simplifies to :
(6+(2/n))/((7/n^3)-11)
I cannot work out how we arrive at this answer.
Help would be greatly appreciated,
kind thanks
i know that 2/n and 7/n^3 tend to zero; given by the reciprocal rule, thus this problem converges to 6/-11 . i get the final answer, its the algebra manipulation beforehand thats bothering me.
thanks
Yes, so one thing you could do is divide the numerator and the demoninator by n^4.
This would leave you with (6 + 2/n)/(7/n^3 - 11)
now as n approaches infinity, the numerator approaches 6. This is because, as n approaches infinity, 2/n goes to 0.
and also, as n approaches infinity, the denominator approaches -11. This is because, as n approaches infinity 7/n^3 also approaches 0.
and hence you are left with 6 in the numerator and -11 in the demoninator, and thus the limit as n approaches infinity is -6/11
oh that wasn't the original post, but if that's what was intended then there you go
thanks, that neat break down really has helped. i'm just stuck with how you reach the last part. i think this is just some basic algebra i'm not quite getting! ie. where does the 2/n and 7/n^3 expressions come from?