Question

已知|x+1|＜ε4，|y-2|＜ε4，|z+3|＜ε4，求证：|x+2y+z|＜ε

已知|x+1|＜ε4，|y-2|＜ε4，|z+3|＜ε4，求证：|x+2y+z|＜ε．

Answer 1

解答：证明：|x+2y+z|=|x+1+2（y-2）+z+3|
≤|x+1|+|2（y-2）|+|z+3|=|x+1|+2|y-2|+|z+3|＜

ε

4

+

ε

2

+

ε

4

=ε．
∴|x+2y+z|＜ε．