How can I show the left and right hand limits of
xsin(1/x) don't exist as x approaches zero?
Thanks
2007-11-02 03:57:38 · 2 answers · asked by mobaxus 2 in Science & Mathematics ➔ Mathematics
I believe they do. Its a hw problem to show they don't exist.
2007-11-02 04:12:17 · update #1
Ooooooops, I there was an extra x in there; it was supposed to be sin(1/x) sorry. And I figured it out, just taking two different subsequences and showing they have different limits.
2007-11-02 07:23:08 · update #2
You can't because they do. The limit is 0 as x-->0.
2007-11-02 04:03:34 · answer #1 · answered by mathematician 7 · 0⤊ 0⤋
Actually, they do exist. the limit of this function as x approaches 0 is 0. The reason is that the sine function remains bounded between -1 and 1, and x approaches 0. To write it out formally, take absolute values and use the "squeeze" theorem.
Are you sure the homework problem is not about left and right hand derivatives? those do not exist.
2007-11-02 04:13:11 · answer #2 · answered by Michael M 7 · 0⤊ 0⤋