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let-f-x-x-x-2-sh-xt-sin-xt-dt-calculate-lim-x-0-f-x-




Question Number 74015 by mathmax by abdo last updated on 17/Nov/19
let f(x) =∫_x ^x^2     ((sh(xt))/(sin(xt)))dt  calculate lim_(x→0) f(x)
letf(x)=xx2sh(xt)sin(xt)dtcalculatelimx0f(x)
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
f(x)=xx2sh(xt)sin(xt)dt=xt=u1xx2x3sh(u)sin(u)duc]x2,x3[/x2x3sh(u)sin(u)du=sh(c)x2x3dusinux2x3dusinu=tan(u2)=ttan(x22)tan(x32)12t1+t2×2dt1+t2=[lnt]tan(x22)tan(x32)=lntan(x32)tan(x22)∣∼lnx(x0)c=λx2+(1λ)x3with0<λ<1f(x)sh(c)xlnxsh(λx2+(1λ)x3)lnx(λx2+(1λ)x3)lnx=x2lnx(λ+(1λ)x)0(x0)limx0f(x)=0

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