搜搜考试网高数上面不懂的问题求解答.
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高数上面不懂的问题求解答.
本人较愚钝.
本人较愚钝.
(1)
lim[(1/x)-1/(e^x-1)]
=lim(e^x-1-x)/[x*(e^x-1)
=lim(e^x-1)/[(e^x-1)+x*e^x]【罗必塔法则】
=lim(e^x-1)/[(x+1)e^x-1]
=lim(e^x)/[e^x+(x+1)*e^x]【再一次】
=lim(e^x)/[(x+2)*e^x]
=lim1/(x+2)
=1/2
(2)
limtan(x³+x²)/ln(x+1)*sinx
=lim[sec²(x³+x²)*(3x²+2x)]/[sinx/(x+1)+ln(x+1)cosx]
=lim(3x²+2x)/[sinx/(x+1)+ln(x+1)]
=lim(3x²+2x)*(x+1)/[sinx+(x+1)ln(x+1)]
=lim(3x³+5x²+2x)/[sinx+(x+1)ln(x+1)]
=lim(9x²+10x+2)/[cosx+ln(x+1)+1]
=(0+0+2)/(1+0+1)
=1
y=x*e^y
===> y'=e^y+x*e^y*y'
===> (1-x*e^y)y'=e^y
===> y'=e^y/(1-x*e^y)
∫xsinxdx
=∫xd(-cosx)
=x*(-cosx)-∫(-cosx)dx
=-xcosx+∫cosxdx
=-xcosx+sinx+C
∫[1/(x+√x)]dx
令√x=t,则x=t²
且x=1时,t=1；x=4时,t=2；且dx=d(t²)=2tdt
原式=∫[2t/(t²+t)]dt
=2∫[1/(t+1)]dt
=2ln(t+1)|
=2(ln3-ln2)
当y=1/x=x时,x=±1
所以,原面积=∫[x-(1/x)]dx
=[(1/2)x²-lnx]|
=[(1/2)*4-ln2]-[(1/2)*1-0]
=(3/2)-ln2
——题目有错!应该是2√x≥3-(1/x)!
令f(x)=2√x+(1/x)-3
定义域为x＞0
则,f'(x)=2*(1/2)*(1/√x)-(1/x²)=1/√x-(1/x²)=[x^(3/2)-1]/(√x*x²)
令f'(x)=0
则,x=1
当0＜x＜1时,f'(x)＜0,f(x)递减；
当x＞1时,f'(x)＞0,f(x)递增.
所以,f(x)在x=1时有最小值f(1=2+1-3=0
所以,x＞0时,f(x)≥0
即,2√x+(1/x)-3≥0
亦即,2√x≥3-(1/x)
lim[(1/x)-1/(e^x-1)]
=lim(e^x-1-x)/[x*(e^x-1)
=lim(e^x-1)/[(e^x-1)+x*e^x]【罗必塔法则】
=lim(e^x-1)/[(x+1)e^x-1]
=lim(e^x)/[e^x+(x+1)*e^x]【再一次】
=lim(e^x)/[(x+2)*e^x]
=lim1/(x+2)
=1/2
(2)
limtan(x³+x²)/ln(x+1)*sinx
=lim[sec²(x³+x²)*(3x²+2x)]/[sinx/(x+1)+ln(x+1)cosx]
=lim(3x²+2x)/[sinx/(x+1)+ln(x+1)]
=lim(3x²+2x)*(x+1)/[sinx+(x+1)ln(x+1)]
=lim(3x³+5x²+2x)/[sinx+(x+1)ln(x+1)]
=lim(9x²+10x+2)/[cosx+ln(x+1)+1]
=(0+0+2)/(1+0+1)
=1
y=x*e^y
===> y'=e^y+x*e^y*y'
===> (1-x*e^y)y'=e^y
===> y'=e^y/(1-x*e^y)
∫xsinxdx
=∫xd(-cosx)
=x*(-cosx)-∫(-cosx)dx
=-xcosx+∫cosxdx
=-xcosx+sinx+C
∫[1/(x+√x)]dx
令√x=t,则x=t²
且x=1时,t=1；x=4时,t=2；且dx=d(t²)=2tdt
原式=∫[2t/(t²+t)]dt
=2∫[1/(t+1)]dt
=2ln(t+1)|
=2(ln3-ln2)
当y=1/x=x时,x=±1
所以,原面积=∫[x-(1/x)]dx
=[(1/2)x²-lnx]|
=[(1/2)*4-ln2]-[(1/2)*1-0]
=(3/2)-ln2
——题目有错!应该是2√x≥3-(1/x)!
令f(x)=2√x+(1/x)-3
定义域为x＞0
则,f'(x)=2*(1/2)*(1/√x)-(1/x²)=1/√x-(1/x²)=[x^(3/2)-1]/(√x*x²)
令f'(x)=0
则,x=1
当0＜x＜1时,f'(x)＜0,f(x)递减；
当x＞1时,f'(x)＞0,f(x)递增.
所以,f(x)在x=1时有最小值f(1=2+1-3=0
所以,x＞0时,f(x)≥0
即,2√x+(1/x)-3≥0
亦即,2√x≥3-(1/x)