4a) $ lim_{t\rightarrow 0} \int_0^1 \frac{e^{-tln(x)}-1}{t} dx = lim_{t\rightarrow 0} \int_0^1 \frac{x^{-t}-1}{t} dx $
$ = lim_{t\rightarrow 0} \frac{1}{t}[\frac{x^{-t+1}}{-t+1}-x]_0^1 = lim_{t\rightarrow 0} \frac{1}{t}[\frac{1^{-t+1}}{-t+1}-1] $
$ = lim_{t\rightarrow 0} \ 0 = 0 $.
4b)
$ lim_{n\rightarrow \infty} \int_1^{n^2} \frac{n cos (\frac{x}{n^2})}{1+nln(x)}dx = lim_{n\rightarrow \infty} \int_1^{\infty} \frac{n cos (\frac{x}{n^2})}{1+nln(x)}\chi_{(1,n^2)} dx \geq \int_1^{\infty} lim_{n\rightarrow \infty} \frac{n cos (\frac{x}{n^2})}{1+nln(x)}\chi_{(1,n^2)} dx $ (Fatou)
$ \geq \int_1^{\infty} lim_{n\rightarrow \infty} \frac{n cos(1)}{1+nln(x)}\chi_{(1,n^2)} dx = \int_1^{\infty} lim_{n\rightarrow \infty} \frac{n cos(1)}{1+nln(x)} dx $
$ = cos(1)\int_1^{\infty} lim_{n\rightarrow \infty} \frac{n}{1+nln(x)} = cos(1)\int_1^{\infty}\frac{1}{ln(x)} \ \forall \ x>1 $ by L'Hospital
and since $ \frac{1}{ln(x)}>\frac{1}{x} \ \forall x>1, cos(1)\int_1^{\infty}\frac{1}{ln(x)}=\infty $
So $ lim_{n\rightarrow \infty} \int_1^{n^2} \frac{n cos (\frac{x}{n^2})}{1+nln(x)}dx =\infty $.
--Wardbc 13:34, 22 July 2008 (EDT)
