Let-f-0-1-R-is-a-continuous-function-prove-that-lim-n-0-1-n-f-x-1-n-2-x-2-dx-pi-2-f-0-proof-S-n-

Question Number 165271 by mnjuly1970 last updated on 28/Jan/22

L e t, f : [0, 1] \to R i s a c o n t i n u o u s

f u n c t i o n, p r o v e t h a t :

{l i m}_{n \to \infty} \int_{0}^{1} \frac{n f (x)}{1 + n^{2} x^{2}} d x = \frac{π}{2} f (0)

- - - p r o o f - - -

S_{n} = [\int_{0}^{\frac{1}{\sqrt{n}}} \frac{n . f (x)}{1 + n^{2} x^{2}} d x = Ω_{n}] + [\int_{\frac{1}{\sqrt{n}}}^{1} \frac{n . f (x)}{1 + n^{2} x^{2}} d x = Φ_{n}]

Ω_{n} \underset{\exists t_{n} \in (0, \frac{1}{\sqrt{n}})}{\overset{M e a n V a l u e T h e o r e m (f i r s t)}{=}} f (t_{n}) \int_{0}^{\frac{1}{\sqrt{n}}} \frac{n}{1 + n^{2} x^{2}} d x

= f (t_{n}) ({t a n}^{- 1} (\sqrt{n}))

{l i m}_{n \to \infty} (Ω_{n}) = \frac{π}{2} f ({l i m}_{n \to \infty} (t_{n})) = \frac{π}{2} f (0)

Φ_{n} = \int_{\frac{1}{\sqrt{n}}}^{1} \frac{n . f (x)}{1 + n^{2} x^{2}} d x \underset{\exists M > 0}{\overset{f i s b o u n d e d}{\Rightarrow}} ∣ Φ_{n} ∣ ⩽ M . \int_{\frac{1}{\sqrt{n}}}^{1} \frac{n}{1 + n^{2} x^{2}} d x

\Rightarrow ∣ Φ_{n} ∣ ⩽ M . ({t a n}^{- 1} (n) - {t a n}^{- 1} (\sqrt{n}))

{l i m}_{n \to \infty} ∣ Φ_{n} ∣ = 0 \Rightarrow {l i m}_{n \to \infty} Φ_{n} = 0

∴ {l i m}_{n \to \infty} (S_{n}) = \frac{π}{2} f (0) m . n

Answered by mindispower last updated on 28/Jan/22

n i c e R e s u l t T h a n x s i r

Commented by mnjuly1970 last updated on 28/Jan/22

t h a n k y o u s o m u c h s i r \dots