(New page: 4a) 4b) <math>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,...)
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Revision as of 13:25, 22 July 2008

4a)


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 $.

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