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posted by  Nelson's Kid on 8/2/2009 10:45:59 AM  |  status: Live  |  Earned Karma: 380

Is the following allowed when taking a limit?

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Say I need to evaluate the limit:   for which f(0) is known to be "1".
Am I allowed to evaluate the function f(h) alone to get   and then use L'Hopitals to find the limit? Thanks
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posted by GUNSH on 8/2/2009 12:26:09 PM  |  status: Live
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 =
                         we know that 
so,
by  Sandwich theorem ,
further ,
GUNSH,you are free to send message if you are stuck anywhere in this solution
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posted by lawrence_tutor on 8/2/2009 12:30:59 PM  |  status: Live
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No.  Your method would imply that the limit always exists, which it doesn't.  Here's a counterexample.

Let f(h) = e^(-xh)(1 + sqrt(h)).  Clearly f(0) = 1.  Also f(h)e^xh - 1 = sqrt(h).  Since lim sqrt(h)/h = lim 1/sqrt(h) does not exist, the given limit does not exist.

Your method would indeed work if f(h) were a factor of the whole numerator, and not just the first term.  Then you could have used the theorem that the limit of a product is the product of the limits to evaluate each limit separately as you suggested.

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posted by lawrence_tutor on 8/2/2009 12:41:39 PM  |  status: Live
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After I posted my result I looked at the previous answer.  Somehow the behaviour of f near 0, which is key to solving the question, disappeared from his answer completely, and rather mysteriously.  Also, the inequality in the second line is wrong.  If you look at the series expansion for e^h - 1, you get h + h^2/2 + h^3/3! + ..., so if you take a positive h, how is this supposed to be <= h?  h is just the first term, and you are adding all those other positive terms as well!



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posted by Nelson's Kid on 8/2/2009 1:12:13 PM  |  status: Live
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Can anyone show me what the actual limit is?
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posted by lawrence_tutor on 8/2/2009 1:56:57 PM  |  status: Live
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Umm, I don't think you understood my solution.  In general there is no limit!  It depends on what f is.  I showed you an example where, for the given f, the limit does not exist.

You are trying to prove a theorem which is not true in general.  Unless you put some more conditions on f, you have no chance to prove anything.

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