Question Number 63273 by mathmax by abdo last updated on 01/Jul/19
$${let}\:{F}\left({x}\right)\:=\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\:\:\:\:\frac{{sin}\left({t}\right)}{{t}+{x}}\:{dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:{F}\left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} {F}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculste}\:{lim}_{{x}\rightarrow\mathrm{0}} \:{F}^{'} \left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} \:{F}^{'} \left({x}\right) \\ $$
Commented by mathmax by abdo last updated on 03/Jul/19
![1) ∃c ∈]x^2 ,x^3 [ wich verify ∫_x^2 ^x^3 ((sint)/(t+x))dt =sinc ∫_x^2 ^x^3 (dt/(t+x)) =sinc [ln∣t+x∣]_(t=x^2 ) ^(t=x^3 ) =sinc {ln∣x^3 +x∣−ln∣x^2 +x∣} =lnc ln∣((x^2 +1)/(x+1))∣ ⇒lim_(x→0) F(x) =lim_(c→0) ×lim_(x→0) ln∣((x^2 +1)/(x+1))∣ =0 we have x^2 ≤t ≤x^3 ⇒x^2 +x ≤t+x≤x+x^3 ⇒for x>0 (1/(x+x^3 ))≤(1/(t+x)) ≤(1/(x^2 +x)) ⇒ (1/(x+x^3 )) ≤∣((sint)/(t+x))∣≤(1/(x^2 +x)) but ∣F(x)∣≤∫_x^2 ^x^3 ∣((sint)/(t+x))∣dt →0 (x→+∞) because lim_(x→+∞) (1/(x+x^3 )) =lim_(x→+∞) (1/(x^2 +x)) =0](Q63396.png)
$$\left.\mathrm{1}\left.\right)\:\exists{c}\:\in\right]{x}^{\mathrm{2}} ,{x}^{\mathrm{3}} \left[\:\:{wich}\:{verify}\:\:\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\:\:\:\frac{{sint}}{{t}+{x}}{dt}\:={sinc}\:\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\frac{{dt}}{{t}+{x}}\right. \\ $$$$={sinc}\:\left[{ln}\mid{t}+{x}\mid\right]_{{t}={x}^{\mathrm{2}} } ^{{t}={x}^{\mathrm{3}} } \:={sinc}\:\left\{{ln}\mid{x}^{\mathrm{3}} \:+{x}\mid−{ln}\mid{x}^{\mathrm{2}} +{x}\mid\right\} \\ $$$$={lnc}\:{ln}\mid\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}+\mathrm{1}}\mid\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{F}\left({x}\right)\:={lim}_{{c}\rightarrow\mathrm{0}} \:\:×{lim}_{{x}\rightarrow\mathrm{0}} {ln}\mid\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}+\mathrm{1}}\mid\:=\mathrm{0} \\ $$$${we}\:{have}\:\:{x}^{\mathrm{2}} \leqslant{t}\:\leqslant{x}^{\mathrm{3}} \:\Rightarrow{x}^{\mathrm{2}} \:+{x}\:\leqslant{t}+{x}\leqslant{x}+{x}^{\mathrm{3}} \:\:\Rightarrow{for}\:{x}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{x}+{x}^{\mathrm{3}} }\leqslant\frac{\mathrm{1}}{{t}+{x}}\:\leqslant\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}}\:\Rightarrow\:\frac{\mathrm{1}}{{x}+{x}^{\mathrm{3}} }\:\leqslant\mid\frac{{sint}}{{t}+{x}}\mid\leqslant\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{x}}\:{but} \\ $$$$\mid{F}\left({x}\right)\mid\leqslant\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \mid\frac{{sint}}{{t}+{x}}\mid{dt}\:\rightarrow\mathrm{0}\:\:\left({x}\rightarrow+\infty\right)\:{because}\:{lim}_{{x}\rightarrow+\infty} \:\frac{\mathrm{1}}{{x}+{x}^{\mathrm{3}} }\:={lim}_{{x}\rightarrow+\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}}\:=\mathrm{0} \\ $$