Recent content by seal308

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'hospital's rule and multiply by conjugate. Was it because...

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 see thanks again.

Wow thanks, that makes so much sense. Ty everyone.

OK I'll use that logic then. thx

Am I overthinking this. Is it as simple as the denominator is larger and gets bigger faster and numerator making the whole thing = 0 for that reason?

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...