华中师范大学：《数学分析》课程PPT教学课件（讲稿）第一章（1.5.2）lim P（x）= lim（anx

讨论若Px)=a0x+axn1+…+an1x+an,则 lim P()=? 提示 imP(x)=lm(ax)+m(a1xn)+…+lm(an21x)+liman x→)x ao lim(xn)+a, lim(xn-)+. +an- lim x+ lim x→>x 0 ao(lim x)n+a,(lim x)n-l+.+an) x→>x0 x→)x =anx0n+a1xn-1+……+an1x0+a P(x0) 上页 下页

上页 返回 下页 •讨论 若 n n P(x)=a0 x n +a1 x n−1 +  +a −1 x+a  则 lim ( ) ? 0 = → P x x x •提示 =a0 x0 n+a1 x0 n−1 +    + an-1x0+an =P(x0 ) n x x n x x n x x n x x x x P x a x a x a x a 0 0 0 0 0 lim ( ) lim ( 0 ) lim ( 1 1 ) lim ( 1 ) lim → − → − → → → = + +  + + n x x x x n n x x n x x a x a x a x a 0 0 0 0 0 lim ( ) 1 lim ( 1 ) 1 lim lim → → − − → → = + +  + + (lim ) (lim ) ) 1 0 1 0 0 n n x x n x x =a x +a x +  +a − → →