An = n^3 / (n + 5) Can I just take the limit and say this is equal to 1 / 0 so it diverges?
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Originally Posted by veronicak5678 An = n^3 / (n + 5) Can I just take the limit and say this is equal to 1 / 0 so it diverges? It diverges because this is equivalent to $\displaystyle \frac{n+125-125}{n+5}=\frac{n^3+125}-\frac{125}{n+5}=n^2+5n+25-\frac{125}{n+5}$ Now what is the limit of that as it goes to infinity?
1 + 0 + 0 - 125/5 = -24 ?
What are you on about? You're subbing 1 here and 0 there. Take the limit as n goes to infinity
Originally Posted by Mathstud28 It diverges because this is equivalent to $\displaystyle \frac{n+125-125}{n+5}=\frac{n^3+125}-\frac{125}{n+5}=n^2+5n+25-\frac{125}{n+5}$ Now what is the limit of that as it goes to infinity? While that's true, isn't is much easier to simply take the original form of A_n and easily see that the limit is infinity as n approaches infinity?
I think that's more along the lines of what I was supposed to be doing, anyway... So I'll just take the limit off the bat and say it diverges.
Originally Posted by veronicak5678 An = n^3 / (n + 5) Can I just take the limit and say this is equal to 1 / 0 so it diverges? $\displaystyle \frac{n^3}{n+5} \geq \frac{n^3}{n+5n}=\tfrac{1}{6}n^2 $ and $\displaystyle \lim ~ \tfrac{1}{6}n^2 = \infty$
Got it. Thanks!
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