例3设f(x)可导,且imf(x)=1求 x→+Q x+2 tsinef(tit x→+oJx 解:由积分中值定理知有∈x,x+21,使 x+2 t sin f(tdt= ssin ∫(5)(x+2-x) x+2 → lim tsinf(nt x→)+Jx 2 lim ssin ef(5)=2 lim 3f($)=6例3 设f(x)可导,且 x lim →+ f (x) = 1 ,求 解: f t dt t t x x x + →+ 2 ( ) 3 lim sin 由积分中值定理知,有[x, x+2],使 f t dt t t x x +2 ( ) 3 sin ( )( 2 ) 3 = sin f x + − x f t dt t t x x x + →+ 2 ( ) 3 lim sin ( ) 3 2 lim sin f →+ = 2 lim 3 ( ) f →+ = =6