?3?P(x)Q(x)R(x)S(x)??
x?a同理，也有h???(a)?3?P(x)Q(x)R(x)S(x)??x?a。因此，g???(a)?h???(a)
13.设函数f(x)在区间[0,??)上连续。若f(x)是非负的增函数，证明函数
?1?F(x)??x?0,?在[0,??)上也是非负的增函数。
?xtf(t)dt,x?00
x?0分析：只需证明函数F(x)在[0,??)上连续，且F?(x)?0(0?x???) 证 首先证明函数F(x)在点0是右连续的，即
xx?0lim?F(x)?lim?x?01x?x0tf(t)dt?xf(x)?0?tf(t)dt?lim?lim?0??0x?0?
x?0?x?0?1其次，当x?0时，
?1F?(x)???x其中
?x0??1tf(t)dt???2x??xtf(t)dt?01xf(x)?xxf(x)?2?2xtf(t)dt0x
?xtf(t)dt?cf(c)0?x，所以 dt?cf(c)x(0?c?x)【积分中值定理】
0x2f(x)?cf(c)xxf(x)?cf(c)[xf(x)?xf(c)]?[xf(c)?cf(c)]?F?(x)?? 2xxxx[f(x)?f(c)]?f(c)(x?c)?0(x?0) ?x因此，F(x)是增函数。
14.求下列极限：
1⑴ limx?0x?0sinx?costdt?lim2x?020sinxcost2dtx?cossin2x?cosx?0???1 ???limx?01?0????⑵ limx??????etdt?2?0?????lim???x??x?2t2edt??022x2?2?2etdt?ex0??2lim22xx??ex?x0etdt22ex????? ????2limexx??2xex?lim1?0 x??x