例7求im(√x+2-√x)(有理化法) x→+Q 解:原式 =lm(√x+2-√x)(√x+2+yx) x→+ x+2+√x x+2-x = x)a√x+2+√x 2 =0 x→+√x+2+yx例7 求 lim ( x 2 x ) x + − →+  解：原式 x x x x x x x + + + − + + = →+  2 ( 2 )( 2 ) lim x x x x x + + + − = →+  2 2 lim x x x + + = →+  2 2 lim =0 (有理化法)