Question Number 165271 by mnjuly1970 last updated on 28/Jan/22
![Let , f : [ 0 , 1 ] → R is a continuous function , prove that : lim_( n→ ∞) ∫_0 ^( 1) (( n f(x))/(1+ n^2 x^( 2) )) dx = (π/2) f (0 ) −−− proof −−− S_( n) = [∫_(0 ) ^( (1/( (√n)))) (( n. f(x))/(1 + n^( 2) x^( 2) )) dx =Ω_( n) ]+[ ∫_(1/( (√n))) ^( 1) ((n.f (x))/(1 + n^( 2) x^( 2) )) dx = Φ_( n) ] Ω_( n) =_(∃ t_( n) ∈ ( 0 , (1/( (√n) )) )) ^(MeanValueTheorem( first)) f (t_( n) )∫_(0 ) ^( (1/( (√n)))) (( n)/(1 + n^( 2) x^( 2) ))dx = f ( t_( n) ) ( tan^( −1) ( (√n) )) lim_( n→∞) (Ω_( n) ) = (π/2) f (lim_( n→∞) ( t_( n) ) ) = (π/2) f (0 ) Φ_( n) = ∫_(1/( (√n))) ^( 1) (( n. f(x) )/(1 + n^( 2) x^( 2) )) dx ⇒_(∃ M >0) ^(f is bounded) ∣ Φ_( n) ∣ ≤ M.∫_(1/( (√n))) ^( 1) (n/(1+ n^( 2) x^( 2) )) dx ⇒ ∣ Φ_( n) ∣ ≤ M . ( tan^( −1) ( n )− tan^( −1) ( (√n) )) lim_( n→ ∞) ∣ Φ_( n) ∣ = 0 ⇒ lim_( n→∞) Φ_( n) =0 ∴ lim_( n→ ∞) ( S_( n) ) = (π/2) f (0 ) ■ m.n](Q165271.png)
$$ \\ $$$$\:\:\:\:\:\mathrm{L}{et}\:,\:\:\:{f}\::\:\left[\:\mathrm{0}\:,\:\mathrm{1}\:\right]\:\rightarrow\:\mathbb{R}\:\:{is}\:{a}\:{continuous}\: \\ $$$$\:\:\:\:{function}\:,\:{prove}\:{that}\::\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{lim}_{\:{n}\rightarrow\:\infty} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{n}\:{f}\left({x}\right)}{\mathrm{1}+\:{n}^{\mathrm{2}} \:{x}^{\:\mathrm{2}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}\:{f}\:\left(\mathrm{0}\:\right) \\ $$$$\:\:\:\:\:−−−\:{proof}\:−−− \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{S}_{\:{n}} \:=\:\left[\int_{\mathrm{0}\:} ^{\:\frac{\mathrm{1}}{\:\sqrt{{n}}}} \:\frac{\:{n}.\:{f}\left({x}\right)}{\mathrm{1}\:+\:{n}^{\:\mathrm{2}} {x}^{\:\mathrm{2}} }\:{dx}\:=\Omega_{\:{n}} \right]+\left[\:\int_{\frac{\mathrm{1}}{\:\sqrt{{n}}}} ^{\:\mathrm{1}} \:\frac{{n}.{f}\:\left({x}\right)}{\mathrm{1}\:+\:{n}^{\:\mathrm{2}} {x}^{\:\mathrm{2}} }\:{dx}\:=\:\Phi_{\:{n}} \:\right] \\ $$$$\:\:\:\:\:\:\Omega_{\:{n}} \:\underset{\exists\:{t}_{\:{n}} \in\:\left(\:\mathrm{0}\:,\:\frac{\mathrm{1}}{\:\sqrt{{n}}\:}\:\right)} {\overset{\mathrm{M}{ean}\mathrm{V}{alue}\mathrm{T}{heorem}\left(\:{first}\right)} {=}}\:\:{f}\:\left({t}_{\:{n}} \:\right)\int_{\mathrm{0}\:} ^{\:\frac{\mathrm{1}}{\:\sqrt{{n}}}} \:\frac{\:{n}}{\mathrm{1}\:+\:{n}^{\:\mathrm{2}} {x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:{f}\:\left(\:{t}_{\:{n}} \right)\:\left(\:{tan}^{\:−\mathrm{1}} \left(\:\sqrt{{n}}\:\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:{lim}_{\:{n}\rightarrow\infty} \:\left(\Omega_{\:{n}} \:\right)\:\:=\:\frac{\pi}{\mathrm{2}}\:{f}\:\left({lim}_{\:{n}\rightarrow\infty} \left(\:{t}_{\:{n}} \right)\:\right)\:=\:\frac{\pi}{\mathrm{2}}\:{f}\:\left(\mathrm{0}\:\right)\: \\ $$$$\:\:\:\:\:\:\:\:\Phi_{\:{n}} =\:\int_{\frac{\mathrm{1}}{\:\sqrt{{n}}}} ^{\:\mathrm{1}} \frac{\:{n}.\:{f}\left({x}\right)\:}{\mathrm{1}\:+\:{n}^{\:\mathrm{2}} {x}^{\:\mathrm{2}} }\:{dx}\:\:\underset{\exists\:\mathrm{M}\:>\mathrm{0}} {\overset{{f}\:\:{is}\:{bounded}} {\Rightarrow}}\:\mid\:\Phi_{\:{n}} \:\mid\:\leqslant\:\mathrm{M}.\int_{\frac{\mathrm{1}}{\:\sqrt{{n}}}} ^{\:\mathrm{1}} \frac{{n}}{\mathrm{1}+\:{n}^{\:\mathrm{2}} {x}^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\mid\:\Phi_{\:{n}} \mid\:\leqslant\:\mathrm{M}\:.\:\left(\:{tan}^{\:−\mathrm{1}} \left(\:{n}\:\right)−\:{tan}^{\:−\mathrm{1}} \left(\:\sqrt{{n}}\:\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:{lim}_{\:{n}\rightarrow\:\infty} \:\mid\:\Phi_{\:{n}} \mid\:=\:\mathrm{0}\:\:\Rightarrow\:{lim}_{\:{n}\rightarrow\infty} \:\Phi_{\:{n}} \:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:{lim}_{\:{n}\rightarrow\:\infty} \:\left(\:\:\mathrm{S}_{\:{n}} \:\right)\:=\:\frac{\pi}{\mathrm{2}}\:{f}\:\left(\mathrm{0}\:\right)\:\:\:\:\blacksquare\:{m}.{n} \\ $$$$\:\:\:\:\: \\ $$
Answered by mindispower last updated on 28/Jan/22
$${nice}\:{Result}\:{Thanx}\:{sir} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 28/Jan/22
$$\:\:\:\:{thank}\:{you}\:{so}\:{much}\:{sir}... \\ $$$$ \\ $$