(1-cos t2) 「例5求极限im。0 x→ 解40利用洛比达法贝 (1-cost)dt 2√x = x→0+ 3 2 lim dl-cos x)=im ix x→>0+5x 0+5x210 2021/2/20
2021/2/20 11 2 5 0 2 0 (1 cos ) [ 5] lim x t dt x x − 例 求极限 → + “ ” ,利用洛比达法则 0 0 2 3 2 5 2 5 2 1 0 0 2 0 (1 cos ) lim (1 cos ) lim x x x t dt x x x x − = − → + → + 2 0 5 (1 cos ) lim x x x − = → + 10 1 5 lim 2 2 2 1 0 = = → + x x x [解]
「例6试问:具有什麽性质的函数,恒有 ∫(x)kx=f()d+C(x∈a,b 若∫∈Ca,b1,则有 ∫fxk=∫f()h+C(x∈,9 2021/2/20 12
2021/2/20 12 [例6] 试 问:具有什麽性质的函数f ,恒 有 f (x)dx f (t)dt C (x [a, b]) x a = + ( ) ( ) ( [ , ]) [ , ], f x dx f t dt C x a b f C a b x a = + 若 则 有