高数极限题,x->0的极限?f=x^2/(sqrt(1+x*sin(x))-sqrt(cos(x))); 4/3 求工程!

高数极限题,x->0的极限?f=x^2/(sqrt(1+x*sin(x))-sqrt(cos(x))); 4/3 求工程!
高数极限题,x->0的极限?f=x^2/(sqrt(1+x*sin(x))-sqrt(cos(x))); 4/3 求工程!

高数极限题,x->0的极限?f=x^2/(sqrt(1+x*sin(x))-sqrt(cos(x))); 4/3 求工程!
f=x^2/(sqrt(1+x*sin(x))-sqrt(cos(x)))
=x^2(sqrt(1+x*sin(x))+sqrt(cos(x))) /(1+x*sin(x)-cos(x))（分母有理化）
(sqrt(1+x*sin(x))+sqrt(cos(x))) 极限为2
考虑g=x^2 /(1+x*sin(x)-cos(x))
罗必达法则：limg=2x/[sinx+xcosx+sinx]=lim2x/[2sinx+xcosx]=lim2/[2cosx+cosx-xsinx]=2/3
或用重要极限：lim1/g=lim[xsinx/x^2+(1-cosx)/x^2]=1+1/2=3/2,故limg=2/3
lmtf=2*2/3=4/3

lim(x→0) x^2/[√(1+xsinx)-√cosx]
x→0, x^2→0, [√(1+xsinx)-√cosx]→0
lim(x→0)x^2/[√(1+xsinx)-√cosx]=lim(x→0)2x/[(1/2)(sinx+xcosx)/√(1+xsinx)+sinx/(2√cosx)]
=lim(x→0)2/[(1/2)(sinx/x+cosx)/√(1+xsinx)+(sinx/x)/(2√cosx)]=2/[(1/2)*2+1/2]=2/(3/2)=4/3