
§1.3极限的运算 一、函数极限的四侧运算定理 如果Iimf(x)=A,limg(x)=B,则 (1)1im[f(x)±g(x)] =limf(x)±limg(x)=A±B. (2)lim[f (x).g(x)] lim f(x).limg(x)=A.B. 3)1im四=imf=-A(此时B≠0. g(x) limg(x)B
§1.3 极限的运算 一、函数极限的四则运算定理 如果lim ( ) f x A = ,lim ( ) g x B = ,则 (1)lim[ ( ) ( )] f x g x = = lim ( ) lim ( ) f x g x A B. (2)lim[ ( ) ( )] f x g x = = lim ( ) lim ( ) f x g x A B. (3) ( ) lim ( ) lim ( ) lim ( ) f x f x A g x g x B = = (此时B 0)

推论1如果Iimf(x)=A,c为常数,则 lim[cf(x)]=clim f(x)=cA. 推论2如果1imf(x)=A,n为常数,则 lim[f(x)]”=[limf(x)]”=A". back next exit
推论 1 如果lim ( ) f x A = ,c为常数,则 lim[ ( )] lim ( ) cf x c f x cA = = . 推论 2 如果lim ( ) f x A = ,n为常数,则 lim[ ( )] [lim ( )] n n n f x f x A = =

例1求1im(3x2-5x+8) x→1 解:1im(3x2-5x+8) x→1 lim3x2-lim5x+lim8 x→1 x→1 x→] =31imx2-51imx+8 X→1 X→ =31imx)2-5+8=3-5+8=6 x→1 A↓人人k水小人小Oaa
例 1 求 2 1 lim(3 5 8) x x x → − + . 解: 2 1 lim(3 5 8) x x x → − + 2 1 1 1 lim3 lim5 lim8 x x x x x → → → = − + . 2 1 1 3lim 5lim 8 x x x x → → = − + . 2 1 3(lim ) 5 8 3 5 8 6 x x → = − + = − + =

若初等函数f(x)在x,处有定义,则有 1imf(x)=f(xo) x→x0 例1另解:1im(3x2-5x+8) x→1 =3×12-5×1+8=6
若初等函数 f x( )在 0 x 处有定义,则有 0 0 lim ( ) ( ) x x f x f x → = . 例 1 另解: 2 1 lim(3 5 8) x x x → − + 2 = − + = 3 1 5 1 8 6

例2求g79 2x 解:因为lim x2-932-9 =0 x→3 2x 2×3 2x 所以1im 5x2-9 三00. back next☐ext
例 2 求 2 3 2 lim x 9 x → x − . 解:因为 2 2 3 9 3 9 lim 0 x 2 2 3 x → x − − = = 所以 2 3 2 lim x 9 x → x = −

例3求1im 2x2-2x+1 x00 5x2+1 2-2 ,1 解:lim 2x2-2x+1=lim-x x X→00 5x2+1 X→0 1 5+ =2-0+0 2 5+0 5
例 3 求 2 2 2 2 1 lim x 5 1 x x → x − + + . 解: 2 2 2 2 1 lim x 5 1 x x → x − + + 2 2 2 1 2 lim 1 5 x x x x → − + = + . 2 0 0 5 0 − + = + . 2 5 =

3x2-2x-1 例4求1im x→2x3-x2+5 321 解:lim 3x2-2x-1 =limxx2x x02x3-x2+5 x→00 15 2-+ 3 0-0-0 =0 2-0+0
例 4 求 2 3 2 3 2 1 lim x 2 5 x x → x x − − − + . 解: 2 3 2 3 2 1 lim x 2 5 x x → x x − − − + 2 3 3 3 2 1 lim 1 5 2 x x x x x x → − − = − + . 000 2 0 0 − − = − + . = 0

例5求1im x3+1 x08x2+2x+9 x3+1 1+ 解:lim 、=lim 8x2+2x+9x8,2,9 =00 一十 2 3 XX back nextext☑
例 5 求 3 2 1 lim x 8 2 9 x → x x + + + . 解: 3 2 1 lim x 8 2 9 x → x x + + + 3 2 3 1 1 limx 8 2 9 x x x x → + = + + . =

由例3~例5得出结论: do n=m 5 lim o n>m x0bx”+bx”+…+bn oon<m Aw↓人人k水人小aaa
由例 3~例 5 得出结论: 0 1 0 0 1 1 0 1 lim 0 m m m n n x n a n m b a x a x a n m b x b x b n m − → − = + + + = + + +

例6求g文-4 x-2 x-2 解:i四产-4 1 = 1110 x→2x+2 4
例 6 求 2 2 2 lim x 4 x → x − − . 解: 2 2 2 lim x 4 x → x − − 2 1 lim x→ x 2 = + . 1 4 =