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Question Number 207648 by SANOGO last updated on 22/May/24

f_{n} (x) = \frac{n}{1 + x^{2}} s i n (\frac{x}{n}) / x \in [0, 1], n ⩾ 1

c a l c u l e r l i m n \to \infty \int_{0}^{1} f_{n} (x) d x

Commented by mr W last updated on 22/May/24

w h y {d o n}^{'} t y o u u s e + \infty i n s t e a d o f

+ o o ?

w h y {d o n}^{'} t y o u u s e \lim_{n \to + \infty} i n s t e a d o f

l i m n \to + o o ?

Answered by Berbere last updated on 22/May/24

l e m m a s i n (x) ⩽ x; \forall x ⩾ 0

s i n (x) = \int_{0}^{x} c o s (t) d t ⩽ \int_{0}^{x} 1 d t = x

f_{n} (x) ⩽ \frac{n}{1 + x^{2}} . \frac{x}{n} = \frac{x}{1 + x^{2}} = G (x)

G (x) i n t e g r a b l e o v e r [0, 1]

\lim_{n \to \infty} \int_{0}^{1} \frac{n}{1 + x^{2}} . s i n (\frac{x}{n}) d x = \int_{0}^{1} \lim_{n \to \infty} . \frac{n s i n (\frac{x}{n})}{1 + x^{2}} d x

= \int_{0}^{1} \frac{s i n (\frac{x}{n}) . x}{\frac{x}{n} (1 + x^{2})} d x; \lim_{x \to 0} \frac{s i n (a)}{a} = 1

= \int_{0}^{1} \frac{x}{1 + x^{2}} d x = {[\frac{l n (1 + x^{2})}{2}]}_{0}^{1} = \frac{l n (2)}{2} = l n (\sqrt{2})

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