So yes you are right, my work will have a negative sign differenece
Consider a function
So now we must find where goes as goes to zero
Which means we must find
Now by letting
Which by the realtion of the Ratio test and Root test is equivalent to
And now seeing that this limit falls under the reign of
The limit is one
some writing ... so i wouldn't be surprised if some typos were found in it! ok, first see that:
now, in your integral, let then:
now since and , we'll have:
it's also clear, by Riemann sum, that:
thus from (2), (3), (4), (5), and (6) we have:
Here is another one I came up with, find L.
Restrictions: You may not use asymptotic equivalences, rate of increase, or in any other way disregard terms.
EDIT: typo, also assuming there is no domain conflictions with other functions