A sequence which tends to 0 but its reciprocal tends to another value
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I have a homework assigned that is asking me to find a sequence $b_n$, whose limit is $0$ but when plugged into the function $f(x) = sin(1/x)$, then $lim_{n to infty}$ $f(b_n) = 1$.
So I thought that I had to find a sequence which does tend to 0 but reciprocal tends to a value of $2/(2k-1)pi$, since that'd when $sin(1/x) = 1$. But so far I'm not able to find such a sequence and would like some tips.
I'm sorry if formating is bad, this is my first post.
analysis
asked Jan 14 at 16:35
Kakano9Kakano9
11
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add a comment |
$begingroup$
I have a homework assigned that is asking me to find a sequence $b_n$, whose limit is $0$ but when plugged into the function $f(x) = sin(1/x)$, then $lim_{n to infty}$ $f(b_n) = 1$.
So I thought that I had to find a sequence which does tend to 0 but reciprocal tends to a value of $2/(2k-1)pi$, since that'd when $sin(1/x) = 1$. But so far I'm not able to find such a sequence and would like some tips.
I'm sorry if formating is bad, this is my first post.
analysis
asked Jan 14 at 16:35
Kakano9Kakano9
11
$endgroup$
$begingroup$
$b_n:=frac{2}{(2n-1)pi}rightarrow 0$ as $nrightarrowinfty$. You solved your own problem, didn't you?
$endgroup$
– Yanko
Jan 14 at 16:38
2
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Your idea looks solid. what's wrong with it? If $b_n=frac 2{(2n-1)pi}$ then $b_nto 0$ but $sin frac 1{b_n}=1$ for all $n$.
$endgroup$
– lulu
Jan 14 at 16:38
$begingroup$
My problem is that since if we take $b_n$ = $2/(2n-1)pi$ then we'll have $1/b_n$ = $(2n-1)pi/2$ but wouldn't the limit of that be infinity
$endgroup$
– Kakano9
Jan 14 at 16:43
1
$begingroup$
you don’t care about $lim 1/b_n$. You only care about $limsin(1/b_n)$. And you did that right.
$endgroup$
– Lubin
Jan 14 at 16:50
add a comment |
$begingroup$
I have a homework assigned that is asking me to find a sequence $b_n$, whose limit is $0$ but when plugged into the function $f(x) = sin(1/x)$, then $lim_{n to infty}$ $f(b_n) = 1$.
So I thought that I had to find a sequence which does tend to 0 but reciprocal tends to a value of $2/(2k-1)pi$, since that'd when $sin(1/x) = 1$. But so far I'm not able to find such a sequence and would like some tips.
I'm sorry if formating is bad, this is my first post.
analysis
asked Jan 14 at 16:35
Kakano9Kakano9
11
$endgroup$
I have a homework assigned that is asking me to find a sequence $b_n$, whose limit is $0$ but when plugged into the function $f(x) = sin(1/x)$, then $lim_{n to infty}$ $f(b_n) = 1$.
So I thought that I had to find a sequence which does tend to 0 but reciprocal tends to a value of $2/(2k-1)pi$, since that'd when $sin(1/x) = 1$. But so far I'm not able to find such a sequence and would like some tips.
I'm sorry if formating is bad, this is my first post.
analysis
analysis
asked Jan 14 at 16:35
Kakano9Kakano9
11
asked Jan 14 at 16:35
Kakano9Kakano9
11
edited Jan 14 at 16:39
Kakano9
asked Jan 14 at 16:35
Kakano9Kakano9
11
asked Jan 14 at 16:35
Kakano9Kakano9
11
asked Jan 14 at 16:35
Kakano9Kakano9
11
11
$begingroup$
$b_n:=frac{2}{(2n-1)pi}rightarrow 0$ as $nrightarrowinfty$. You solved your own problem, didn't you?
$endgroup$
– Yanko
Jan 14 at 16:38
2
$begingroup$
Your idea looks solid. what's wrong with it? If $b_n=frac 2{(2n-1)pi}$ then $b_nto 0$ but $sin frac 1{b_n}=1$ for all $n$.
$endgroup$
– lulu
Jan 14 at 16:38
$begingroup$
My problem is that since if we take $b_n$ = $2/(2n-1)pi$ then we'll have $1/b_n$ = $(2n-1)pi/2$ but wouldn't the limit of that be infinity
$endgroup$
– Kakano9
Jan 14 at 16:43
1
$begingroup$
you don’t care about $lim 1/b_n$. You only care about $limsin(1/b_n)$. And you did that right.
$endgroup$
– Lubin
Jan 14 at 16:50
add a comment |
$begingroup$
$b_n:=frac{2}{(2n-1)pi}rightarrow 0$ as $nrightarrowinfty$. You solved your own problem, didn't you?
$endgroup$
– Yanko
Jan 14 at 16:38
2
$begingroup$
Your idea looks solid. what's wrong with it? If $b_n=frac 2{(2n-1)pi}$ then $b_nto 0$ but $sin frac 1{b_n}=1$ for all $n$.
$endgroup$
– lulu
Jan 14 at 16:38
$begingroup$
My problem is that since if we take $b_n$ = $2/(2n-1)pi$ then we'll have $1/b_n$ = $(2n-1)pi/2$ but wouldn't the limit of that be infinity
$endgroup$
– Kakano9
Jan 14 at 16:43
1
$begingroup$
you don’t care about $lim 1/b_n$. You only care about $limsin(1/b_n)$. And you did that right.
$endgroup$
– Lubin
Jan 14 at 16:50
$begingroup$
$b_n:=frac{2}{(2n-1)pi}rightarrow 0$ as $nrightarrowinfty$. You solved your own problem, didn't you?
$endgroup$
– Yanko
Jan 14 at 16:38
$begingroup$
$b_n:=frac{2}{(2n-1)pi}rightarrow 0$ as $nrightarrowinfty$. You solved your own problem, didn't you?
$endgroup$
– Yanko
Jan 14 at 16:38
2
2
$begingroup$
Your idea looks solid. what's wrong with it? If $b_n=frac 2{(2n-1)pi}$ then $b_nto 0$ but $sin frac 1{b_n}=1$ for all $n$.
$endgroup$
– lulu
Jan 14 at 16:38
$begingroup$
Your idea looks solid. what's wrong with it? If $b_n=frac 2{(2n-1)pi}$ then $b_nto 0$ but $sin frac 1{b_n}=1$ for all $n$.
$endgroup$
– lulu
Jan 14 at 16:38
$begingroup$
My problem is that since if we take $b_n$ = $2/(2n-1)pi$ then we'll have $1/b_n$ = $(2n-1)pi/2$ but wouldn't the limit of that be infinity
$endgroup$
– Kakano9
Jan 14 at 16:43
$begingroup$
My problem is that since if we take $b_n$ = $2/(2n-1)pi$ then we'll have $1/b_n$ = $(2n-1)pi/2$ but wouldn't the limit of that be infinity
$endgroup$
– Kakano9
Jan 14 at 16:43
1
1
$begingroup$
you don’t care about $lim 1/b_n$. You only care about $limsin(1/b_n)$. And you did that right.
$endgroup$
– Lubin
Jan 14 at 16:50
$begingroup$
you don’t care about $lim 1/b_n$. You only care about $limsin(1/b_n)$. And you did that right.
$endgroup$
– Lubin
Jan 14 at 16:50
add a comment |
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$begingroup$
$b_n:=frac{2}{(2n-1)pi}rightarrow 0$ as $nrightarrowinfty$. You solved your own problem, didn't you?
$endgroup$
– Yanko
Jan 14 at 16:38
2
$begingroup$
Your idea looks solid. what's wrong with it? If $b_n=frac 2{(2n-1)pi}$ then $b_nto 0$ but $sin frac 1{b_n}=1$ for all $n$.
$endgroup$
– lulu
Jan 14 at 16:38
$begingroup$
My problem is that since if we take $b_n$ = $2/(2n-1)pi$ then we'll have $1/b_n$ = $(2n-1)pi/2$ but wouldn't the limit of that be infinity
$endgroup$
– Kakano9
Jan 14 at 16:43
1
$begingroup$
you don’t care about $lim 1/b_n$. You only care about $limsin(1/b_n)$. And you did that right.
$endgroup$
– Lubin
Jan 14 at 16:50