发布网友 发布时间:2024-10-03 01:50
共1个回答
热心网友 时间:2024-12-01 15:43
x=0时原式=0,
x>0时原式=lim<n→+∞>[(1-n^(-2x)]/[1+n^(-2x)]=1,
仿上,x<0,原式=-1.
f(x-1)
=lim(n→+∞)((n+x)/n)^n
=lim(n→+∞)(1+x/n)^n
=lim(n→+∞)(1+x/n)^(n/x·x)
=[lim(n→+∞)(1+x/n)^(n/x)]^x
=e^x
令t=x-1,则x=t+1
所以,f(t)=e^(t+1)
所以,f(x)=e^(x+1)
令lnx=t,则x=e^t, 且当x→+∞时,t→+∞
于是原式=lim【t→+∞】t^n/e^t..............∞/∞型,用洛比达法则
= lim【t→+∞】nt^(n-1)/e^t......若n-1>0,则继续用洛比达法则,直至t的指数变为0或者负数
于是可知原式=0
不明白可以追问,如果有帮助,请选为满意回答!
lim(x->+∞)时,sinx+cosx是有限值,先不计
则lim(x->+∞),(x^3+x^2+1)/(2^x+x^3)
=lim(x->+∞),(1+1/x+1/x^3)/(2^x/x^3+1)
=lim(x->+∞),1/(2^x/x^3+1)
又lim(x->+∞),(2^x/x^3)经过多次洛必达法则知其极限为+∞
所以原极限为0
lim{f(a+1/n)/f(a)}^n
=lim{[1+[f(a+1/n)-f(a)]/f(a)]^{f(a)/[f(a+1/n)-f(a)]}^[f(a+1/n)-f(a)]/[1/nf(a)]}
由于lim{[1+[f(a+1/n)-f(a)]/f(a)]^{f(a)/[f(a+1/n)-f(a)]}
是重要极限,其值为e
而lim[f(a+1/n)-f(a)]/[1/nf(a)]
可利用f(x)在x=a处可导,的定义
=lim[f(a+1/n)-f(a)]/[1/n]·[1/f(a)]
=f'(a)/f(a)
所以,原极限值为:
e^(f'(a)/f(a))
lim(x->∞) (√(x^2+x)-√(x^2+1))
=lim(x->∞) ( x-1) /(√(x^2+x)+√(x^2+1))
=lim(x->∞) ( 1-1/x) /(√(1+1/x)+√(1+1/x))
=1/2
=1
lim<n→∞>cos(x/2)…cos(x/2^n) (x≠0)
= lim<n→∞>cos(x/2)…cos(x/2^n)2sin(x/2^n)/[2sin(x/2^n)]
= lim<n→∞>cos(x/2)…cos[x/2^(n-1)]sin[x/2^(n-1)]/[2sin(x/2^n)]
= lim<n→∞>cos(x/2)…cos[x/2^(n-2)]sin[x/2^(n-)]/[2^2sin(x/2^n)]
=............
= lim<n→∞>cos(x/2)sin(x/2)/[2^(n-1)sin(x/2^n)]
= lim<n→∞>sinx/[2^nsin(x/2^n)]
= lim<n→∞>sinx/[2^n(x/2^n)] = sinx/x
lim x趋近于正无穷 (e^x-1)/x (用罗比达法则)
=lim x趋近于正无穷 e^x=正无穷
因cos x /2cosx/4…cosx/2^n
=[cosx/2*cosx/4*....*2sinx/2^n*cosx/2^n]/(2sinx/2^n)
=[cosx/2*cosx/4*...*sinx/2^(n-1)]/(2sinx/2^n)
=(cosx/2sinx/2)/[2^(n-1)*sin(x/2^n]
=sinx/[2^n*sin(x/2^n)]
所以lim (n趋近正无穷) cos x /2cosx/4…cosx/2^n
=lim (n趋近正无穷) sinx/[x*sin(x/2^n)/(x/2^n)]
=(sinx)/x