lim x趋向于π/4 求 (tanx)^tan2xlim x趋向于0 求(x^2 * csc x * sin 1/x)已知 x1=1,x2=1+x1/(x1+1),Xn=1+1/(1+Xn-1) 求数列Xn的极限

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limx趋向于π/4求(tanx)^tan2xlimx趋向于0求(x^2*cscx*sin1/x)已知x1=1,x2=1+x1/(x1+1),Xn=1+1/(1+Xn-1)求数列Xn的极限limx趋向

lim x趋向于π/4 求 (tanx)^tan2xlim x趋向于0 求(x^2 * csc x * sin 1/x)已知 x1=1,x2=1+x1/(x1+1),Xn=1+1/(1+Xn-1) 求数列Xn的极限
lim x趋向于π/4 求 (tanx)^tan2x
lim x趋向于0 求(x^2 * csc x * sin 1/x)
已知 x1=1,x2=1+x1/(x1+1),Xn=1+1/(1+Xn-1) 求数列Xn的极限

lim x趋向于π/4 求 (tanx)^tan2xlim x趋向于0 求(x^2 * csc x * sin 1/x)已知 x1=1,x2=1+x1/(x1+1),Xn=1+1/(1+Xn-1) 求数列Xn的极限
1.原式=lim(x->π/4)[(tanx)^tan(2x)]
=e^{lim(x->π/4)[ln(tanx)/tan(2x)]}
=e^{lim(x->π/4)[cos²(2x)/sin(2x)]} (应用罗比达法则,并化简)
=e^(0/1)
=1
2.原式=lim(x->0)[x²cscx*sin(1/x)]
=lim(x->0){(x/sinx)*[xsin(1/x)]}
=[lim(x->0)(x/sinx)]*{lim(x->0)[xsin(1/x)]}
=1*0 (应用特殊极限lim(x->0)(sinx/x)=1)
=0
3.∵X1=1,X2=1+1/(X1+1),Xn=1+1/(1+Xn-1)
∴X2=3/2,00,Xn+1=F(Xn-1)
∵X2n+3-X2n+1=F(X2n+1)-F(X2n-1)=F′(c)(X2n+1-X2n-1) (X2n+1∞)Xn=A
∴由Xn=1+1/(1+Xn-1)两边取极限得:
A=1+1/(1+A)==>A+A²=1+A+1
==>A²=2
==>A=±√2
∵Xn>0,则A>0
∴A=√2
故数列{Xn}的极限是√2.