Let be
And since
can be written as
but the numerator is also equal to
Then sub and into and respectively , obtaining
and
therefore
Problem :
Is it related to INTEGRAL ? If you think it is not , i still have another problem
maybe you will be confused about what i have said,here is a proof
let (it exists)
and the only restriction is that
now we will show
so, for some
but
thus
let x->inf (sup inf), then
let epsilon->0, then
similarily,
so
The proof is copyed from my book. Usually we require numerator to ->inf, but it is unnecessary.
Problem: Let be continuous and suppose that exists and it's finite. Prove that [/quote]
The solution is here.
l'Hôpital's rule - Wikipedia, the free encyclopedia
http://upload.wikimedia.org/math/5/f...c513fb6556.png
Check last paragraph in Section "Examples".
is the Heaviside or unit step function, see here:
Heaviside Step Function -- from Wolfram MathWorld