?e20. lim??sin2x?3x2x?0xlim?e?sin2x3x2?limx?0x?0xxlim?e?2.
?11?. ??x?0xln?1?x???1?1ln?1?x??xln?1?x??x11?x?lim?lim??【解析】原式?lim. 2x?0xln?x?0x?01?x?2x2xtanx?sinx. 3x?0xsinx12?sinxx1?cosx1cosx2【解析】原式=lim. ?lim?lim?x?0x?0x2cosxx?0x2cosxx3221. lim22.limx?01?xsinx?1e?1x2.
112xsinxx1【解析】原式=lim22?lim22?.
x?0x?0x2x23.(5分) lim?x.
x?0lnx1x1x1x2xxlnx?ex?0【解析】原式?lim?ex?0limxlnx?x?0?limx?0?lim?e?e??e0?1.
x3?ax2?x?424.设lim 具有极限l,求a,l的值.
x??1x?1【解析】因为lim(x?1)?0,所以 lim(x?ax?x?4)?0,
x??1x??132因此 a?4 并将其代入原式
x3?4x2?x?4(x?1)(x?1)(x?4)l?lim?lim?10
x??1x??1x?1x?125.若limf(x)存在,且f(x)?x?3x?2limf(x),求f(x)和limf(x).
x?1x?1x?126 / 7
2【解析】设limf(x)?A,对等式f(x)?x?3x?2limf(x)两边同时取极限x?1可得,
x?1limf(x)?limx2?3x?2limf(x),即A?limx2?3x?2A,故limf(x)?A??4.
x?1x?12?x?1?x?1??x?1??x?1所以f(x)?x?3x?8. 四.证明题
26.设函数f(x),g(x)均在闭区间?a,b?上连续,且有f(a)?g(a)?a,f(b)?g(b)?b,证明:存在??,使f(a,b)????g?????成立.
【证明】 构造函数F(x)?f(x)?g(x)?x,则函数F(x)在闭区间?a,b?上连续, 而F(a)?f(a)?g(a)?a?0,F(b)?f(b)?g(b)?b?0, 显然F(a)?F(b)?0
于是由连续函数的零点定理知,??(a,b),使得F(?)?0, 即 存在??,使f(a,b)
????g?????.
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