61、数学竞赛辅导之-极限1

1?122?2elim1?x?ex?02x?x22?e?2elimx?02x(1?x)??e?e22?08.求极限limn→∞n(n-1)xn解:

x???limx(x?1)=lime1lnxxx-1x1-21xx→+∞1x=lim-112?limlnxx1?2xlnx?lim=0x???xx→+∞x???nnn9.求极限lim(2+22+?22)n→∞n+1n+2n+n解：原式

1=limn→∞n12∑(i/n)i=11+n=∫0112dx1+x10=arctanx|?=4(0)≠0，求lim10.设f′(x)连续,f(0)=0,f′?xx20f(t)dtxx?02?0f(t)dt解:原式=limx→0?lim2xf(x)2xf(t)dt+xf(x)∫02x22f(x)2?f(t)dt?xf(x)022x?0x4xf′(x)=limx→02f(x)+f(x)+xf′(x)2′4f(x)=limf(x)x→03+f′(x)x2′4f(x)=limf(x)-f(0)x→03+f′(x)x-02′4f(x)=limx→03f′(0)+f′(0)=1