Can someone show me how:

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- Dec 8th 2008, 02:21 AMGreengoblinlimit = e
Can someone show me how:

- Dec 8th 2008, 02:26 AMJameson
Say that this limit exists and it equals A.

Then

The LHS is also and when evaluating at t=0 you get 0/0, which is indeterminate and L'Hopitals can be used.

So

Try the limit again, and use algebra to solve for A again.

edit: This is in the precalc section, so forgive me for using calculus. - Dec 8th 2008, 02:30 AMChop Suey
- Dec 8th 2008, 02:35 AMJameson
Yes that would appear so after looking some more :)

- Dec 8th 2008, 03:57 AMGreengoblin
Thank you both so much, I understand it now. :)

- Dec 8th 2008, 09:24 AMJhevon
Alternate solution:

recall that for any .

so that:

now since (By L'Hopital's rule, or power series or ...)

we have that the limit is

This is very similar to Jameson's method, both of these methods you should consider when taking the limit of an expression with the variable in the power.

That is, let and be expressions in a single variable in which the variable appears, then we can usually find various limits that give indeterminate forms by doing the following:

The method Jameson used:

Say

Then

so that

and so by taking the anti-log of both sides.

The method I used:

simply write as , and take the limit of both sides.

note that - Dec 8th 2008, 09:43 AMGreengoblin
Thanks, but could you explain how ?

- Dec 8th 2008, 09:45 AMChop Suey
If and is continuous on , then

- Dec 8th 2008, 09:48 AMJhevon
the rigorous theory might get a bit complicated, but it is somewhat similar to the fact that we can factor constants out side of limits. since the limit deals with X and not e, taking the limit of e to some power really only affects the power, since anything that is not a variable in X is not affected by the limit.

EDIT: Chop Suey was a bit more rigorous :p. - Dec 9th 2008, 02:59 AMGreengoblin
Thanks, I got it now. ;) I need to have a good look at limits again, as I haven't looked at them for ages and have forgotten alot.