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Limits question

limit as n tends to infinity

((e^-n+n)/n+2))^2n

I thought if I found the limit of the bracket to be 1 and 1^2n is always 1 therefore the limit is 1?
Original post by CammieInfinity
limit as n tends to infinity

((e^-n+n)/n+2))^2n

I thought if I found the limit of the bracket to be 1 and 1^2n is always 1 therefore the limit is 1?


Try a large number for n on your calculator. It isn't 1 is it?
I can't remember much about limits but this can be turned into:

lim⁡x→∞((1+1x)x)−4\lim_{x\to \infty} ((1+\frac{1}{x})^x)^{-4} where n=2xn=2x

Your notation is ambiguous by the way - you could do with a bracket around n + 2.
(edited 9 years ago)


How did you get it to 1+1/x? The division yields (1+(e^(-n)-2/n+2))^2n
maths limits.png97.0KB
Original post by CammieInfinity
How did you get it to 1+1/x? The division yields (1+(e^(-n)-2/n+2))^2n


If you still can't be bothered to use unambiguous notation I can't be bothered to make the effort to try to read that. I used the fact that e−xe^{-x} tends to zero as x tends to infinity.
I got it despite your short comings

https://www.youtube.com/watch?v=wrZRrzfMF3o
Original post by CammieInfinity
...


Don't start being unpleasant or your time here could be very short.
Oh no say it isn't so, whatever will I do without your guidance. Plus i have several e-mails and signing up takes two ticks - you're fighting a lost battle hun. Xoxo

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