Question Number 207648 by SANOGO last updated on 22/May/24
![f_n (x)=(n/(1+x^2 ))sin((x/n)) /x∈[0,1] , n≥1 calculer lim n→∞∫_0 ^1 f_n (x)dx](https://www.tinkutara.com/question/Q207648.png)
$${f}_{{n}} \left({x}\right)=\frac{{n}}{\mathrm{1}+{x}^{\mathrm{2}} }{sin}\left(\frac{{x}}{{n}}\right)\:\:/{x}\in\left[\mathrm{0},\mathrm{1}\right]\:,\:\:{n}\geqslant\mathrm{1} \\ $$$${calculer}\:{lim}\:{n}\rightarrow\infty\int_{\mathrm{0}} ^{\mathrm{1}} {f}_{{n}} \left({x}\right){dx} \\ $$
Commented by mr W last updated on 22/May/24
$${why}\:{don}'{t}\:{you}\:{use}\:+\infty\:{instead}\:{of}\: \\ $$$$+{oo}? \\ $$$${why}\:{don}'{t}\:{you}\:{use}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{instead}\:{of}\: \\ $$$${lim}\:{n}\rightarrow+{oo}? \\ $$
Answered by Berbere last updated on 22/May/24
![lemma sin(x)≤x;∀x≥0 sin(x)=∫_0 ^x cos(t)dt≤∫_0 ^x 1dt=x f_n (x)≤(n/(1+x^2 )).(x/n)=(x/(1+x^2 ))=G(x) G(x) integrable over [0,1] lim_(n→∞) ∫_0 ^1 (n/(1+x^2 )).sin((x/n))dx=∫_0 ^1 lim_(n→∞) .((nsin((x/n)))/(1+x^2 ))dx =∫_0 ^1 ((sin((x/n)).x)/((x/n)(1+x^2 )))dx;lim_(x→0) ((sin(a))/a)=1 =∫_0 ^1 (x/(1+x^2 ))dx=[((ln(1+x^2 ))/2)]_0 ^1 =((ln(2))/2)=ln((√2))](https://www.tinkutara.com/question/Q207659.png)
$${lemma}\:{sin}\left({x}\right)\leqslant{x};\forall{x}\geqslant\mathrm{0} \\ $$$${sin}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {cos}\left({t}\right){dt}\leqslant\int_{\mathrm{0}} ^{{x}} \mathrm{1}{dt}={x} \\ $$$${f}_{{n}} \left({x}\right)\leqslant\frac{{n}}{\mathrm{1}+{x}^{\mathrm{2}} }.\frac{{x}}{{n}}=\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }={G}\left({x}\right) \\ $$$${G}\left({x}\right)\:{integrable}\:{over}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{n}}{\mathrm{1}+{x}^{\mathrm{2}} }.{sin}\left(\frac{{x}}{{n}}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\rightarrow\infty} {\mathrm{lim}}.\frac{{nsin}\left(\frac{{x}}{{n}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left(\frac{{x}}{{n}}\right).{x}}{\frac{{x}}{{n}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx};\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\left({a}\right)}{{a}}=\mathrm{1} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\left[\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}={ln}\left(\sqrt{\mathrm{2}}\right) \\ $$