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lim x→无穷 (sin 2/x +Cos 1/x)的x次方

一楼的过程是错误的.“故相于求极限 x*ln(sin2/x+cos1/x) 由于x->无穷时,cos1/x->1 故上式造价于x*ln(sin2/x+1)”过程有问题.将x*ln(sin2/x+cos1/x)变为x*ln(sin2/x+1)有问题.y=(sin 2/x +cos 1/x)^x lny=xln(sin 2/x +cos 1/x).limlnx=lim[xln(sin 2/x +

lim(x→∞) [cos(1/x)+sin(2/x)]^(1/x),这样真的好难打,还不如发图=lim(x→∞) [cos(1/x)+2sin(1/x)cos(1/x)]^(1/x)=lim(x→∞) {cos(1/x)[1+2sin(1/x)cos(1/x)]}^(1/x)=lim(x→∞) [cos(1/x)]^(1/x)*lim(x→∞) [1+2sin(1/x)]^(1/x)=1*lim(x→∞) [1+2sin(1/x)]^{1/[2sin(1/x)]}*[2sin(1/x)]/x=e^2lim(x→∞) sin(1/x)/x=e^0=1

J(x) = [sin(2/x)+cos(1/x)]^(x)lnJ(x) = ln[sin(2/x)+cos(1/x)]/(1/x) 当 x->∞ 时,lnJ(x) = 0/0 不定式,采用洛必达法则lim(x->∞) lnJ(x)= lim(x->∞) [-2(1/x^2)cos(2/x)+(1/x^2)sin(1/x)]/{[sin(2/x)+cos(1/x)](-1/x^2)}= lim(x->∞) [2cos(2/x)-sin(1/x)]/[sin(2/x)+cos(1/x)]= 2得到: lim(x->∞) lnJ(x) = 2即 : lim(x->∞) J(x) = e^2 (近似等于 ≈ 7.3890)

lim(sin1/x∧2+cos1/x)^(x2) (x→∞) 对原式取对数,若原式极限为E1,取完之后极限为E2,则lnE1=E2,E1=e^E2 lim ln((sin1/x∧2+cos1/x)^(x2) ) (x→∞) =lim (x2) ln(sin1/x∧2+cos1/x) (x→∞) =lim ln(sin1/x∧2+cos1/x)/(1/x2) (x→∞) =lim ln(sint∧2+cost)/(t2

原式=e^(xln(sin1/x+cos1/x-1+1)ln(1+x)~x x0ln(sin1/x+cos1/x-1+1)~sin1/x+cos1/x-1x(sin1/x+cos1/x-1)=x(sin1/x-2sin(1/2x)^2)=2xsin(1/2x)(cos1/2x -sin1/2x)=2x*1/2x=1sin1/2x~1/2x值为e

lim (sin(2/x)+cos(1/x))^x (x→∞)sin(2/x)→0 cos(1/x)→∞1=1

答:lim(x→∞) [cos(2/x)+sin(1/x)]^x=lim(t→0) (cos2t+sint)^(1/t)=lim(t→0) (1+t)^(1/t)=e

解:令1/x=t,则x→∞时t→0. 容易求得 lim(t→0)ln(sin2t+cost)/t =lim(t→0)(ln(sin2t+cost))'/(t)' (洛必达法则) =lim(t→0)(2cos2t-sint)/(sin2t+cost) =(2cos0-sin0)/(sin0+cos0) =2 因此 原式=lim(t→0)(sin2t+cost)^(1/t) =e^2

题目应该是当x逼近到0得时候,limx^2*cos(1/x)=0lim(sin(x^2*cos(1/x)))/x=lim(x^2*cos(1/x))/x=lim(x*cos(1/x))=0

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