Show as $xrightarrow 0+, cos(1/x) / x < infty$ [closed]
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I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.
My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.
real-analysis limits measure-theory
asked Dec 3 '18 at 1:20
rakaraka
556
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closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.
My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.
real-analysis limits measure-theory
asked Dec 3 '18 at 1:20
rakaraka
556
$endgroup$
closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
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– RRL
Dec 3 '18 at 1:33
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Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
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– raka
Dec 3 '18 at 1:35
3
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It is not bounded near $0$
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– RRL
Dec 3 '18 at 1:37
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I see it now. Thank you.
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– raka
Dec 3 '18 at 1:38
add a comment |
$begingroup$
I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.
My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.
real-analysis limits measure-theory
asked Dec 3 '18 at 1:20
rakaraka
556
$endgroup$
I am trying to show that given
$$f(x) = xsin(1/x),quad x>0,quad f(0)=0, $$
the derivative is bounded. Namely, $ f'(x) < infty$
for $ x in (0,a)$ for some $ainmathbb{R}$.
My attempt at a solution lead me to taking the derivative using the Product Rule but I can't find a way to justify that $-frac{1}{x}cos(1/x) < infty$
as $x rightarrow 0^+$ despite Wolfram-Alpha telling me that the limit is indeed $< infty$.
real-analysis limits measure-theory
real-analysis limits measure-theory
asked Dec 3 '18 at 1:20
rakaraka
556
asked Dec 3 '18 at 1:20
rakaraka
556
edited Dec 3 '18 at 1:35
raka
asked Dec 3 '18 at 1:20
rakaraka
556
asked Dec 3 '18 at 1:20
rakaraka
556
asked Dec 3 '18 at 1:20
rakaraka
556
556
closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, RRL, José Carlos Santos, Leucippus, Cesareo Dec 3 '18 at 10:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
$endgroup$
– RRL
Dec 3 '18 at 1:33
$begingroup$
Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
$endgroup$
– raka
Dec 3 '18 at 1:35
3
$begingroup$
It is not bounded near $0$
$endgroup$
– RRL
Dec 3 '18 at 1:37
$begingroup$
I see it now. Thank you.
$endgroup$
– raka
Dec 3 '18 at 1:38
add a comment |
$begingroup$
WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
$endgroup$
– RRL
Dec 3 '18 at 1:33
$begingroup$
Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
$endgroup$
– raka
Dec 3 '18 at 1:35
3
$begingroup$
It is not bounded near $0$
$endgroup$
– RRL
Dec 3 '18 at 1:37
$begingroup$
I see it now. Thank you.
$endgroup$
– raka
Dec 3 '18 at 1:38
$begingroup$
WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
$endgroup$
– RRL
Dec 3 '18 at 1:33
$begingroup$
WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
$endgroup$
– RRL
Dec 3 '18 at 1:33
$begingroup$
Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
$endgroup$
– raka
Dec 3 '18 at 1:35
$begingroup$
Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
$endgroup$
– raka
Dec 3 '18 at 1:35
3
3
$begingroup$
It is not bounded near $0$
$endgroup$
– RRL
Dec 3 '18 at 1:37
$begingroup$
It is not bounded near $0$
$endgroup$
– RRL
Dec 3 '18 at 1:37
$begingroup$
I see it now. Thank you.
$endgroup$
– raka
Dec 3 '18 at 1:38
$begingroup$
I see it now. Thank you.
$endgroup$
– raka
Dec 3 '18 at 1:38
add a comment |
1 Answer
1
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votes
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Take $x={1over{2pi n}}$, $f'(x)=-2pi n$
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $x={1over{2pi n}}$, $f'(x)=-2pi n$
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
add a comment |
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Take $x={1over{2pi n}}$, $f'(x)=-2pi n$
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
add a comment |
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Take $x={1over{2pi n}}$, $f'(x)=-2pi n$
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
Take $x={1over{2pi n}}$, $f'(x)=-2pi n$
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
answered Dec 3 '18 at 1:27
Tsemo AristideTsemo Aristide
57k11444
57k11444
add a comment |
add a comment |
$begingroup$
WolframAlpha tells you the limit is undefined in the interval $(-infty,+infty)$ which means there are sequences $x_n, y_n $ converging to $0+$ such that $f(x_n) to +infty$ and $f(y_n) to -infty$. It definitely does not mean a limit "$< infty$" exists
$endgroup$
– RRL
Dec 3 '18 at 1:33
$begingroup$
Right. I misused the word limit. All I really want to show is that the derivative is bounded in $mathbb{R}$. I adjusted the notation.
$endgroup$
– raka
Dec 3 '18 at 1:35
3
$begingroup$
It is not bounded near $0$
$endgroup$
– RRL
Dec 3 '18 at 1:37
$begingroup$
I see it now. Thank you.
$endgroup$
– raka
Dec 3 '18 at 1:38