设{an}中,an=(3n+1)/(4n-1),则
|an-3/4|=|(3n+1)/(4n-1)-3/4|=|7/[4*(4n-1)]|解得n>7/(16E)+1/4,
所以取N=[7/(16E)+1/4]("[]"为取整函数),
则任意取E>0,总存在N∈Z+,当n＞N时,有
|an-3/4|lim=3/4