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Don't understand convergence as n approaches infinity |
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| Oct11-12, 07:15 PM | #1 |
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Don't understand convergence as n approaches infinity
Here's the deal....
I don't understand the limit as n→∞ of [(1+(.05/n))^20n -1]/[.05/n] My Calculus book says that it's supposed to approach {e^[(.05)(20)]-1}/[.05], but the numerator is a constant while the denominator goes to 0 as n→∞. The textbook, by Dr. Gilbert Strang, has similar limits, and its not in the selected errata, so I doubt it was just some error, can someone help me? Just about any light anyone could shed on this would be helpful :] Cheers, Mathguy |
| Oct11-12, 09:55 PM | #2 |
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This looks like an indeterminate form of the type "(1^(infinity) - 1)/0", from which it is not clear whether it converges or not. You cannot be sure that it is a nonzero constant over 0, as it might be that the numerator's limit is actually 0 (and you'd get a 0/0 form).
To find this limit, you can't use direct substitution, you must use something like l'Hopital's Rule, probably twice or more (some algebraic manipulations might be required before applying l'Hopital's Rule). |
| Oct11-12, 10:03 PM | #3 |
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But the numerator is (1+(.05/n))^20n)-1, which is e^[(.05)(20)]-1=e-1 in the limit, so it has to be a constant (unless I missed something here. I just want to understand it D:)
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| Oct12-12, 11:38 PM | #4 |
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Recognitions:
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Don't understand convergence as n approaches infinity
That looks like a typo, what page of what edition is that on?
[(1+(.05/n))^(20n) -1]/[.05]->{e^[(.05)(20)]-1}/[.05]=20(e-1) [(1+(.05/n))^(20n) -1]/[.05/n]->infinity |
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