Determine the limit of the function $f$ at $0$, for the function defined on $mathbb{R}$ by $f(0)=0$ and...
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I have a problem of analysis, for which I'd like to ask some tips how to begin, precisely I need to determine the limit of the function, if it's exists, by using a formal definition of a limit (epsilon delta).
The function is defined by the following:
$f(0) = 0 $ if $x = 0$
$f(x) = sin(1/x)$ if $x neq 0 $
Domain of the definition of the function is $mathbb{R}$
real-analysis limits functions continuity
edited Jan 20 at 22:47
rtybase
11k21533
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
$endgroup$
add a comment |
$begingroup$
I have a problem of analysis, for which I'd like to ask some tips how to begin, precisely I need to determine the limit of the function, if it's exists, by using a formal definition of a limit (epsilon delta).
The function is defined by the following:
$f(0) = 0 $ if $x = 0$
$f(x) = sin(1/x)$ if $x neq 0 $
Domain of the definition of the function is $mathbb{R}$
real-analysis limits functions continuity
edited Jan 20 at 22:47
rtybase
11k21533
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
$endgroup$
$begingroup$
Consider $x_n=frac{1}{left(2n+frac 12right)pi}.$ Get $f(x_n).$
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– mfl
Jan 20 at 14:27
$begingroup$
You could try using a sequence converging to $0$, e.g.$x_n = 1/n$ and see what happens with $f(x_n)$, just to forge yourself some opinion.
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– J.F
Jan 20 at 14:28
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maybe you mean $xsin(1/x)$ instead of $sin(1/x)$; this has a limit at zero.
$endgroup$
– Karl
Jan 20 at 14:40
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@Karl ,I meant sin(1/x). If it has a limit at 0, what method, in general I should use to prove that using a formal definition of limit ?
$endgroup$
– AverageMarcin
Jan 20 at 14:47
add a comment |
$begingroup$
I have a problem of analysis, for which I'd like to ask some tips how to begin, precisely I need to determine the limit of the function, if it's exists, by using a formal definition of a limit (epsilon delta).
The function is defined by the following:
$f(0) = 0 $ if $x = 0$
$f(x) = sin(1/x)$ if $x neq 0 $
Domain of the definition of the function is $mathbb{R}$
real-analysis limits functions continuity
edited Jan 20 at 22:47
rtybase
11k21533
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
$endgroup$
I have a problem of analysis, for which I'd like to ask some tips how to begin, precisely I need to determine the limit of the function, if it's exists, by using a formal definition of a limit (epsilon delta).
The function is defined by the following:
$f(0) = 0 $ if $x = 0$
$f(x) = sin(1/x)$ if $x neq 0 $
Domain of the definition of the function is $mathbb{R}$
real-analysis limits functions continuity
real-analysis limits functions continuity
edited Jan 20 at 22:47
rtybase
11k21533
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
edited Jan 20 at 22:47
rtybase
11k21533
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
edited Jan 20 at 22:47
rtybase
11k21533
edited Jan 20 at 22:47
rtybase
11k21533
edited Jan 20 at 22:47
rtybase
11k21533
11k21533
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
asked Jan 20 at 14:24
AverageMarcinAverageMarcin
84
84
$begingroup$
Consider $x_n=frac{1}{left(2n+frac 12right)pi}.$ Get $f(x_n).$
$endgroup$
– mfl
Jan 20 at 14:27
$begingroup$
You could try using a sequence converging to $0$, e.g.$x_n = 1/n$ and see what happens with $f(x_n)$, just to forge yourself some opinion.
$endgroup$
– J.F
Jan 20 at 14:28
$begingroup$
maybe you mean $xsin(1/x)$ instead of $sin(1/x)$; this has a limit at zero.
$endgroup$
– Karl
Jan 20 at 14:40
$begingroup$
@Karl ,I meant sin(1/x). If it has a limit at 0, what method, in general I should use to prove that using a formal definition of limit ?
$endgroup$
– AverageMarcin
Jan 20 at 14:47
add a comment |
$begingroup$
Consider $x_n=frac{1}{left(2n+frac 12right)pi}.$ Get $f(x_n).$
$endgroup$
– mfl
Jan 20 at 14:27
$begingroup$
You could try using a sequence converging to $0$, e.g.$x_n = 1/n$ and see what happens with $f(x_n)$, just to forge yourself some opinion.
$endgroup$
– J.F
Jan 20 at 14:28
$begingroup$
maybe you mean $xsin(1/x)$ instead of $sin(1/x)$; this has a limit at zero.
$endgroup$
– Karl
Jan 20 at 14:40
$begingroup$
@Karl ,I meant sin(1/x). If it has a limit at 0, what method, in general I should use to prove that using a formal definition of limit ?
$endgroup$
– AverageMarcin
Jan 20 at 14:47
$begingroup$
Consider $x_n=frac{1}{left(2n+frac 12right)pi}.$ Get $f(x_n).$
$endgroup$
– mfl
Jan 20 at 14:27
$begingroup$
Consider $x_n=frac{1}{left(2n+frac 12right)pi}.$ Get $f(x_n).$
$endgroup$
– mfl
Jan 20 at 14:27
$begingroup$
You could try using a sequence converging to $0$, e.g.$x_n = 1/n$ and see what happens with $f(x_n)$, just to forge yourself some opinion.
$endgroup$
– J.F
Jan 20 at 14:28
$begingroup$
You could try using a sequence converging to $0$, e.g.$x_n = 1/n$ and see what happens with $f(x_n)$, just to forge yourself some opinion.
$endgroup$
– J.F
Jan 20 at 14:28
$begingroup$
maybe you mean $xsin(1/x)$ instead of $sin(1/x)$; this has a limit at zero.
$endgroup$
– Karl
Jan 20 at 14:40
$begingroup$
maybe you mean $xsin(1/x)$ instead of $sin(1/x)$; this has a limit at zero.
$endgroup$
– Karl
Jan 20 at 14:40
$begingroup$
@Karl ,I meant sin(1/x). If it has a limit at 0, what method, in general I should use to prove that using a formal definition of limit ?
$endgroup$
– AverageMarcin
Jan 20 at 14:47
$begingroup$
@Karl ,I meant sin(1/x). If it has a limit at 0, what method, in general I should use to prove that using a formal definition of limit ?
$endgroup$
– AverageMarcin
Jan 20 at 14:47
add a comment |
1 Answer
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We have that the sequence $$x_n=frac{1}{left(2n+frac 12right)pi}$$ converges to $0$ as $nto infty.$ But
$$f(x_n)=sin left(2n+frac 12right)pi=1, forall ninmathbb N.$$
Also the sequence $$y_n=frac{1}{2npi}$$ converges to $0$ as $nto infty.$ But
$$f(y_n)=sin 2npi=0, forall ninmathbb N.$$
Thus, the limit doesn't exist.
answered Jan 20 at 15:46
mflmfl
26.5k12141
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add a comment |
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1 Answer
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$begingroup$
We have that the sequence $$x_n=frac{1}{left(2n+frac 12right)pi}$$ converges to $0$ as $nto infty.$ But
$$f(x_n)=sin left(2n+frac 12right)pi=1, forall ninmathbb N.$$
Also the sequence $$y_n=frac{1}{2npi}$$ converges to $0$ as $nto infty.$ But
$$f(y_n)=sin 2npi=0, forall ninmathbb N.$$
Thus, the limit doesn't exist.
answered Jan 20 at 15:46
mflmfl
26.5k12141
$endgroup$
add a comment |
$begingroup$
We have that the sequence $$x_n=frac{1}{left(2n+frac 12right)pi}$$ converges to $0$ as $nto infty.$ But
$$f(x_n)=sin left(2n+frac 12right)pi=1, forall ninmathbb N.$$
Also the sequence $$y_n=frac{1}{2npi}$$ converges to $0$ as $nto infty.$ But
$$f(y_n)=sin 2npi=0, forall ninmathbb N.$$
Thus, the limit doesn't exist.
answered Jan 20 at 15:46
mflmfl
26.5k12141
$endgroup$
add a comment |
$begingroup$
We have that the sequence $$x_n=frac{1}{left(2n+frac 12right)pi}$$ converges to $0$ as $nto infty.$ But
$$f(x_n)=sin left(2n+frac 12right)pi=1, forall ninmathbb N.$$
Also the sequence $$y_n=frac{1}{2npi}$$ converges to $0$ as $nto infty.$ But
$$f(y_n)=sin 2npi=0, forall ninmathbb N.$$
Thus, the limit doesn't exist.
answered Jan 20 at 15:46
mflmfl
26.5k12141
$endgroup$
We have that the sequence $$x_n=frac{1}{left(2n+frac 12right)pi}$$ converges to $0$ as $nto infty.$ But
$$f(x_n)=sin left(2n+frac 12right)pi=1, forall ninmathbb N.$$
Also the sequence $$y_n=frac{1}{2npi}$$ converges to $0$ as $nto infty.$ But
$$f(y_n)=sin 2npi=0, forall ninmathbb N.$$
Thus, the limit doesn't exist.
answered Jan 20 at 15:46
mflmfl
26.5k12141
answered Jan 20 at 15:46
mflmfl
26.5k12141
answered Jan 20 at 15:46
mflmfl
26.5k12141
answered Jan 20 at 15:46
mflmfl
26.5k12141
26.5k12141
add a comment |
add a comment |
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$begingroup$
Consider $x_n=frac{1}{left(2n+frac 12right)pi}.$ Get $f(x_n).$
$endgroup$
– mfl
Jan 20 at 14:27
$begingroup$
You could try using a sequence converging to $0$, e.g.$x_n = 1/n$ and see what happens with $f(x_n)$, just to forge yourself some opinion.
$endgroup$
– J.F
Jan 20 at 14:28
$begingroup$
maybe you mean $xsin(1/x)$ instead of $sin(1/x)$; this has a limit at zero.
$endgroup$
– Karl
Jan 20 at 14:40
$begingroup$
@Karl ,I meant sin(1/x). If it has a limit at 0, what method, in general I should use to prove that using a formal definition of limit ?
$endgroup$
– AverageMarcin
Jan 20 at 14:47