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Question Number 74015 by mathmax by abdo last updated on 17/Nov/19
$${let}\:{f}\left({x}\right)\:=\int_{{x}} ^{{x}^{\mathrm{2}} } \:\:\:\frac{{sh}\left({xt}\right)}{{sin}\left({xt}\right)}{dt} \\ $$$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right) \\ $$
Commented by mathmax by abdo last updated on 18/Nov/19
![f(x)=∫_x ^x^2 ((sh(xt))/(sin(xt)))dt =_(xt =u) (1/x) ∫_x^2 ^x^3 ((sh(u))/(sin(u)))du ∃c ∈]x^2 ,x^3 [ / ∫_x^2 ^x^3 ((sh(u))/(sin(u)))du =sh(c) ∫_x^2 ^x^3 (du/(sinu)) ∫_x^2 ^x^3 (du/(sinu)) =_(tan((u/2))=t) ∫_(tan((x^2 /2))) ^(tan((x^3 /2))) (1/((2t)/(1+t^2 )))×((2dt)/(1+t^2 )) =[ln∣t∣]_(tan((x^2 /2))) ^(tan(((x3)/2))) =ln∣((tan((x^3 /2)))/(tan((x^2 /2))))∣∼ln∣x∣ (x→0) c=λx^2 +(1−λ)x^3 with 0<λ<1 ⇒f(x)∼((sh(c))/x)ln∣x∣ ∼sh(λx^2 +(1−λ)x^3 )ln∣x∣ ∼ (λx^2 +(1−λ)x^3 )ln∣x∣ =x^2 ln∣x∣(λ +(1−λ)x)→0 (x→0) ⇒lim_(x→0) f(x)=0](Q74086.png)
$${f}\left({x}\right)=\int_{{x}} ^{{x}^{\mathrm{2}} } \:\:\frac{{sh}\left({xt}\right)}{{sin}\left({xt}\right)}{dt}\:=_{{xt}\:={u}} \:\:\frac{\mathrm{1}}{{x}}\:\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\frac{{sh}\left({u}\right)}{{sin}\left({u}\right)}{du} \\ $$$$\left.\exists{c}\:\in\right]{x}^{\mathrm{2}} ,{x}^{\mathrm{3}} \left[\:/\:\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\frac{{sh}\left({u}\right)}{{sin}\left({u}\right)}{du}\:={sh}\left({c}\right)\:\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\:\frac{{du}}{{sinu}}\right. \\ $$$$\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\frac{{du}}{{sinu}}\:=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)={t}} \:\:\:\:\int_{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\left[{ln}\mid{t}\mid\right]_{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$={ln}\mid\frac{{tan}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\right)}{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}\mid\sim{ln}\mid{x}\mid\:\:\:\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${c}=\lambda{x}^{\mathrm{2}} \:+\left(\mathrm{1}−\lambda\right){x}^{\mathrm{3}} \:\:\:{with}\:\mathrm{0}<\lambda<\mathrm{1}\:\Rightarrow{f}\left({x}\right)\sim\frac{{sh}\left({c}\right)}{{x}}{ln}\mid{x}\mid \\ $$$$\sim{sh}\left(\lambda{x}^{\mathrm{2}} \:+\left(\mathrm{1}−\lambda\right){x}^{\mathrm{3}} \right){ln}\mid{x}\mid\:\sim\:\:\left(\lambda{x}^{\mathrm{2}} \:+\left(\mathrm{1}−\lambda\right){x}^{\mathrm{3}} \right){ln}\mid{x}\mid \\ $$$$={x}^{\mathrm{2}} {ln}\mid{x}\mid\left(\lambda\:+\left(\mathrm{1}−\lambda\right){x}\right)\rightarrow\mathrm{0}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{f}\left({x}\right)=\mathrm{0} \\ $$$$ \\ $$