Can someone show me how:
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.
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
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 .