Paradoxica
-insert title here-
lol he didn't mention it that was me inserting my commentary into his quote.If you need to estimate the integrand anyway, I would just go ahead and use the Taylor bounds you mentioned.
=\frac{1}{x}\int_x^{4x} \cos(1/u)\, du\Rightarrow \\ \\ 3=\frac{1}{x}\int_x^{4x}\, du\geq F(x)\geq \frac{1}{x}\int_x^{4x}(1-\frac{1}{2u^2})\, du = 3-\frac{3}{8x^2}\\ \\ \Rightarrow \lim_{x\rightarrow\infty} F(x)=3. )
Conditions on L'Hopital's rule (1 Viewer)
- Thread starter leehuan
- Start date
seanieg89
Well-Known Member
- Joined
- Aug 8, 2006
- Messages
- 2,653
- Gender
- Male
- HSC
- 2007
Lol, it always catches me off guard when people edit the things they quote.
0/0, since numerator goes to ln(1) = 0.I was basically prepared to bash this with L'Hopital's rule just like the answers did, but what "indeterminate form" is this supposed to be again?
}}{x}})