I similified the denominator as follows:
sqrt(n^2 + 3n -4) + sqrt(n^2 + 6n +5)
which becomes (n^2)^(1/2) (1 + 3/n + 4/n^2)^1/2 + (n^2)^(1/2) (1+6/n + 5/n)^1/2
(n^2)^(1/2) simplifies to just n, factor n out. and 3/n, 4/n^2, 6/n and 5/n all go to 0 as n approaches infinity so i just said they...
I get it!
Simplify denom is 2n so do limit as n approaches infinity of (-3n - 9) /(2n)
Factor out the n: limit as n approaches infinity of n(-3 - 9/n) /n(2)
n cancels out, 9/n goes to 0 so you get -3/2.
But how do you know not to use l'hopitals rule and multiply by conjugate.
Was it because it...
Hi,
I need help with the following limit, the solution is apparently -3/2 but I don't get it.
Question:
limit as n approaches infinity of [ sqrt(n^2 +3n - 4) - sqrt(n^2 + 6n +5) ]
Attempt: So I was just thinking to factor out n
like: (n^2)^(1/2) (1 + 3/n - 4/n^2)^1/2 - n (n^2)^(1/2)...
I'm just learnign about sequences right now.
What does making the limit inside mean?
Do you mean something like it becomes lim(3^(n+1) / lim(4^(n+1))
and because the denominator is bigger than the numerator it goes to zero?
Hello,
I was just wondering how to solve this limit: Limit of (3/4)^(n+1) as n approaches infinity
My attempt:
(3/4)^(n+1) = (3^ (n+1) ) / (4 ^ (n+1))
Top goes to infinity and bottom goes to infinity to use l'hopital rule.
lim = ( ln(3) * 1 * 3^(n+1) ) / ( ln(4) * 1 * 4^(n+1) )
But the...
Hello,
I was just wondering how to solve this limit: Limit of (3/4)^(n+1) as n approaches infinity
My attempt:
(3/4)^(n+1) = (3^ (n+1) ) / (4 ^ (n+1))
Top goes to infinity and bottom goes to infinity to use l'hopital rule.
lim = ( ln(3) * 1 * 3^(n+1) ) / ( ln(4) * 1 * 4^(n+1) )
But the...