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求极限:lim(n→∞)2^nsinx/2^n(x不为零的常数);lim(x→0)(tanx-sinx)/x^3 ;lim(x→a)(sinx-sina)/(x-a
lim(x→π/3)sin(x-π/3)/(1-2cosx) 希望写写过程,
人气:119 ℃ 时间:2019-10-23 04:30:52
解答
①等价无穷小量替换:
lim(n→∞)2^nsin(x/2^n)
=lim(n→∞)2^n*(x/2^n)
= x

【罗必塔法则】
lim(x→0)(tanx-sinx)/x^3
=lim(x→0)(sec^2 x - cosx)/3x^2
=lim(x→0)(2sec^2 xtanx + sinx)/6x
=lim(x→0)(2sec^2 x/cosx + 1)*sinx/6x
= 3*(1/6)
= 1/2

lim(x->a) [sinx - sina]/[x-a]
=lim(x->a) { 2cos[(x+a)/2][sin(x-a)/2] }/ [x-a]
=lim(x->a)cos[(x+a)/2]* {2[(x-a)/2}/[x-a]
=cosa

【罗必塔法则】
lim(x→π/3)sin(x-π/3)/(1-2cosx)
=lim(x→π/3) cos(x-π/3)/2sinx
= 1/(2*1/2)
= 1
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