How can I find a sequence $a_n$ whose limit tends to $0$, but also for which $sinfrac{1}{a_n}$ tends to $0$?
$begingroup$
Am I correct in thinking that I need $frac{1}{a_n}$ to tend to a multiple of $pi$ so that $sin$ of this equals $0$ since a sequence tending to $0$ and its reciprocal tending to $0$ would be impossible. If so, how can I find such a sequence as I am really struggling?
real-analysis limits
edited Jan 16 at 22:49
whiskeyo
1368
asked Jan 16 at 22:10
DanDan
11
$endgroup$
add a comment |
$begingroup$
Am I correct in thinking that I need $frac{1}{a_n}$ to tend to a multiple of $pi$ so that $sin$ of this equals $0$ since a sequence tending to $0$ and its reciprocal tending to $0$ would be impossible. If so, how can I find such a sequence as I am really struggling?
real-analysis limits
edited Jan 16 at 22:49
whiskeyo
1368
asked Jan 16 at 22:10
DanDan
11
$endgroup$
6
$begingroup$
Try something like $a_n=frac 1{pi,n}$.
$endgroup$
– lulu
Jan 16 at 22:13
add a comment |
$begingroup$
Am I correct in thinking that I need $frac{1}{a_n}$ to tend to a multiple of $pi$ so that $sin$ of this equals $0$ since a sequence tending to $0$ and its reciprocal tending to $0$ would be impossible. If so, how can I find such a sequence as I am really struggling?
real-analysis limits
edited Jan 16 at 22:49
whiskeyo
1368
asked Jan 16 at 22:10
DanDan
11
$endgroup$
Am I correct in thinking that I need $frac{1}{a_n}$ to tend to a multiple of $pi$ so that $sin$ of this equals $0$ since a sequence tending to $0$ and its reciprocal tending to $0$ would be impossible. If so, how can I find such a sequence as I am really struggling?
real-analysis limits
real-analysis limits
edited Jan 16 at 22:49
whiskeyo
1368
asked Jan 16 at 22:10
DanDan
11
edited Jan 16 at 22:49
whiskeyo
1368
asked Jan 16 at 22:10
DanDan
11
edited Jan 16 at 22:49
whiskeyo
1368
edited Jan 16 at 22:49
whiskeyo
1368
edited Jan 16 at 22:49
whiskeyo
1368
1368
asked Jan 16 at 22:10
DanDan
11
asked Jan 16 at 22:10
DanDan
11
asked Jan 16 at 22:10
DanDan
11
11
6
$begingroup$
Try something like $a_n=frac 1{pi,n}$.
$endgroup$
– lulu
Jan 16 at 22:13
add a comment |
6
$begingroup$
Try something like $a_n=frac 1{pi,n}$.
$endgroup$
– lulu
Jan 16 at 22:13
6
6
$begingroup$
Try something like $a_n=frac 1{pi,n}$.
$endgroup$
– lulu
Jan 16 at 22:13
$begingroup$
Try something like $a_n=frac 1{pi,n}$.
$endgroup$
– lulu
Jan 16 at 22:13
add a comment |
0
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$begingroup$
Try something like $a_n=frac 1{pi,n}$.
$endgroup$
– lulu
Jan 16 at 22:13