Mixing
Limit of a Function that is the Product of two Functions.
$begingroup$
I am trying to show the following,
Let $f : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$lim_{x to infty} f(x) = lim_{x to 0} f(x) = 0$$
Let $g : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$$
Show that $lim_{x to infty} g(x) = lim_{x to 0} g(x) = 0$.
My attempt:
We know $g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$ and $$ 0 leq left[1+text{sin}left(2 pi text{ln}(x)right)right] leq 2$$
Hence, $$lim_{x to infty} g(x) = lim_{x to infty} f(x) = 0$$
and $$lim_{x to 0} g(x) = lim_{x to 0} f(x) = 0$$
Is this correct / rigorous enough? Is there a better way of doing this? I am not sure that the fact that $1 + text{sin}(2 pi text{ln}(x))$ is bounded is enough, since some limits don't converge due to rapid oscillation. How can I make my attempt better?
Does this principle apply in general when we have some limiting behaviour of a function $f$, and a bounded function $g$, is the limiting behaviour of $fg$ the same as that of $f$?
calculus limits functions proof-verification
asked May 22 '18 at 1:19
E-muE-mu
807417
$endgroup$
add a comment |
$begingroup$
I am trying to show the following,
Let $f : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$lim_{x to infty} f(x) = lim_{x to 0} f(x) = 0$$
Let $g : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$$
Show that $lim_{x to infty} g(x) = lim_{x to 0} g(x) = 0$.
My attempt:
We know $g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$ and $$ 0 leq left[1+text{sin}left(2 pi text{ln}(x)right)right] leq 2$$
Hence, $$lim_{x to infty} g(x) = lim_{x to infty} f(x) = 0$$
and $$lim_{x to 0} g(x) = lim_{x to 0} f(x) = 0$$
Is this correct / rigorous enough? Is there a better way of doing this? I am not sure that the fact that $1 + text{sin}(2 pi text{ln}(x))$ is bounded is enough, since some limits don't converge due to rapid oscillation. How can I make my attempt better?
Does this principle apply in general when we have some limiting behaviour of a function $f$, and a bounded function $g$, is the limiting behaviour of $fg$ the same as that of $f$?
calculus limits functions proof-verification
asked May 22 '18 at 1:19
E-muE-mu
807417
$endgroup$
2
$begingroup$
This is correct, although I would highlight that $g$ is the product of $f$ by something that is bounded.
$endgroup$
– Fimpellizieri
May 22 '18 at 1:29
1
$begingroup$
You can make more explicit what you are using. For example, the inequalities that you wrote imply $0leq |g(x)|=|f(x)||1+sin(2piln(x))|leq 2|f(x)|$. Then, by the whichever-way-you-call-it theorem, the limit of $|g(x)|$ is zero. This implies that the limit of $g(x)$ is also zero.
$endgroup$
– user561348
May 22 '18 at 1:29
$begingroup$
$g(x) = f(x)h(x).$ If $lim_limits{xto a} f(x), lim_limits{xto a} h(x)$ exist $lim_limits{xto a} g(x) = (lim_limits{xto a} f(x) )(lim_limits{xto a} h(x)), h(x)$ is bounded. i.e. $0le h(x)le 2, 0<lim_limits{xto a} g(x) <2lim_limits{xto a} f(x)$
$endgroup$
– Doug M
May 22 '18 at 1:36
add a comment |
$begingroup$
I am trying to show the following,
Let $f : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$lim_{x to infty} f(x) = lim_{x to 0} f(x) = 0$$
Let $g : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$$
Show that $lim_{x to infty} g(x) = lim_{x to 0} g(x) = 0$.
My attempt:
We know $g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$ and $$ 0 leq left[1+text{sin}left(2 pi text{ln}(x)right)right] leq 2$$
Hence, $$lim_{x to infty} g(x) = lim_{x to infty} f(x) = 0$$
and $$lim_{x to 0} g(x) = lim_{x to 0} f(x) = 0$$
Is this correct / rigorous enough? Is there a better way of doing this? I am not sure that the fact that $1 + text{sin}(2 pi text{ln}(x))$ is bounded is enough, since some limits don't converge due to rapid oscillation. How can I make my attempt better?
Does this principle apply in general when we have some limiting behaviour of a function $f$, and a bounded function $g$, is the limiting behaviour of $fg$ the same as that of $f$?
calculus limits functions proof-verification
asked May 22 '18 at 1:19
E-muE-mu
807417
$endgroup$
I am trying to show the following,
Let $f : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$lim_{x to infty} f(x) = lim_{x to 0} f(x) = 0$$
Let $g : mathbb{R^+} to mathbb{R}$ be a real-valued function such that $$g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$$
Show that $lim_{x to infty} g(x) = lim_{x to 0} g(x) = 0$.
My attempt:
We know $g(x) = f(x)left[1+text{sin}left(2 pi text{ln}(x)right)right]$ and $$ 0 leq left[1+text{sin}left(2 pi text{ln}(x)right)right] leq 2$$
Hence, $$lim_{x to infty} g(x) = lim_{x to infty} f(x) = 0$$
and $$lim_{x to 0} g(x) = lim_{x to 0} f(x) = 0$$
Is this correct / rigorous enough? Is there a better way of doing this? I am not sure that the fact that $1 + text{sin}(2 pi text{ln}(x))$ is bounded is enough, since some limits don't converge due to rapid oscillation. How can I make my attempt better?
Does this principle apply in general when we have some limiting behaviour of a function $f$, and a bounded function $g$, is the limiting behaviour of $fg$ the same as that of $f$?
calculus limits functions proof-verification
calculus limits functions proof-verification
asked May 22 '18 at 1:19
E-muE-mu
807417
asked May 22 '18 at 1:19
E-muE-mu
807417
edited Jan 24 at 15:27
E-mu
asked May 22 '18 at 1:19
E-muE-mu
807417
asked May 22 '18 at 1:19
E-muE-mu
807417
asked May 22 '18 at 1:19
E-muE-mu
807417
807417
2
$begingroup$
This is correct, although I would highlight that $g$ is the product of $f$ by something that is bounded.
$endgroup$
– Fimpellizieri
May 22 '18 at 1:29
1
$begingroup$
You can make more explicit what you are using. For example, the inequalities that you wrote imply $0leq |g(x)|=|f(x)||1+sin(2piln(x))|leq 2|f(x)|$. Then, by the whichever-way-you-call-it theorem, the limit of $|g(x)|$ is zero. This implies that the limit of $g(x)$ is also zero.
$endgroup$
– user561348
May 22 '18 at 1:29
$begingroup$
$g(x) = f(x)h(x).$ If $lim_limits{xto a} f(x), lim_limits{xto a} h(x)$ exist $lim_limits{xto a} g(x) = (lim_limits{xto a} f(x) )(lim_limits{xto a} h(x)), h(x)$ is bounded. i.e. $0le h(x)le 2, 0<lim_limits{xto a} g(x) <2lim_limits{xto a} f(x)$
$endgroup$
– Doug M
May 22 '18 at 1:36
add a comment |
2
$begingroup$
This is correct, although I would highlight that $g$ is the product of $f$ by something that is bounded.
$endgroup$
– Fimpellizieri
May 22 '18 at 1:29
1
$begingroup$
You can make more explicit what you are using. For example, the inequalities that you wrote imply $0leq |g(x)|=|f(x)||1+sin(2piln(x))|leq 2|f(x)|$. Then, by the whichever-way-you-call-it theorem, the limit of $|g(x)|$ is zero. This implies that the limit of $g(x)$ is also zero.
$endgroup$
– user561348
May 22 '18 at 1:29
$begingroup$
$g(x) = f(x)h(x).$ If $lim_limits{xto a} f(x), lim_limits{xto a} h(x)$ exist $lim_limits{xto a} g(x) = (lim_limits{xto a} f(x) )(lim_limits{xto a} h(x)), h(x)$ is bounded. i.e. $0le h(x)le 2, 0<lim_limits{xto a} g(x) <2lim_limits{xto a} f(x)$
$endgroup$
– Doug M
May 22 '18 at 1:36
2
2
$begingroup$
This is correct, although I would highlight that $g$ is the product of $f$ by something that is bounded.
$endgroup$
– Fimpellizieri
May 22 '18 at 1:29
$begingroup$
This is correct, although I would highlight that $g$ is the product of $f$ by something that is bounded.
$endgroup$
– Fimpellizieri
May 22 '18 at 1:29
1
1
$begingroup$
You can make more explicit what you are using. For example, the inequalities that you wrote imply $0leq |g(x)|=|f(x)||1+sin(2piln(x))|leq 2|f(x)|$. Then, by the whichever-way-you-call-it theorem, the limit of $|g(x)|$ is zero. This implies that the limit of $g(x)$ is also zero.
$endgroup$
– user561348
May 22 '18 at 1:29
$begingroup$
You can make more explicit what you are using. For example, the inequalities that you wrote imply $0leq |g(x)|=|f(x)||1+sin(2piln(x))|leq 2|f(x)|$. Then, by the whichever-way-you-call-it theorem, the limit of $|g(x)|$ is zero. This implies that the limit of $g(x)$ is also zero.
$endgroup$
– user561348
May 22 '18 at 1:29
$begingroup$
$g(x) = f(x)h(x).$ If $lim_limits{xto a} f(x), lim_limits{xto a} h(x)$ exist $lim_limits{xto a} g(x) = (lim_limits{xto a} f(x) )(lim_limits{xto a} h(x)), h(x)$ is bounded. i.e. $0le h(x)le 2, 0<lim_limits{xto a} g(x) <2lim_limits{xto a} f(x)$
$endgroup$
– Doug M
May 22 '18 at 1:36
$begingroup$
$g(x) = f(x)h(x).$ If $lim_limits{xto a} f(x), lim_limits{xto a} h(x)$ exist $lim_limits{xto a} g(x) = (lim_limits{xto a} f(x) )(lim_limits{xto a} h(x)), h(x)$ is bounded. i.e. $0le h(x)le 2, 0<lim_limits{xto a} g(x) <2lim_limits{xto a} f(x)$
$endgroup$
– Doug M
May 22 '18 at 1:36
add a comment |
1 Answer
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$begingroup$
You should clarify the use of the squeeze theorem: since $0le 1+sin(2piln x)le 2$, we also have
$$
0le g(x)le 2f(x)
$$
and since $lim_{xto?}f(x)=0$ (where $?=infty$ or $?=0^+$), we can conclude that
$$
lim_{xto?}g(x)=0
$$
This uses substantially that the limits of $f$ are both zero, otherwise the application of the squeeze theorem cannot be done. A simple example: $f(x)=1$ and $g(x)=f(x)(1+sin(2piln x))$. Neither the limit at $0^+$ nor the limit at $infty$ exist.
answered Jan 24 at 15:51
egregegreg
184k1486205
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add a comment |
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$begingroup$
You should clarify the use of the squeeze theorem: since $0le 1+sin(2piln x)le 2$, we also have
$$
0le g(x)le 2f(x)
$$
and since $lim_{xto?}f(x)=0$ (where $?=infty$ or $?=0^+$), we can conclude that
$$
lim_{xto?}g(x)=0
$$
This uses substantially that the limits of $f$ are both zero, otherwise the application of the squeeze theorem cannot be done. A simple example: $f(x)=1$ and $g(x)=f(x)(1+sin(2piln x))$. Neither the limit at $0^+$ nor the limit at $infty$ exist.
answered Jan 24 at 15:51
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You should clarify the use of the squeeze theorem: since $0le 1+sin(2piln x)le 2$, we also have
$$
0le g(x)le 2f(x)
$$
and since $lim_{xto?}f(x)=0$ (where $?=infty$ or $?=0^+$), we can conclude that
$$
lim_{xto?}g(x)=0
$$
This uses substantially that the limits of $f$ are both zero, otherwise the application of the squeeze theorem cannot be done. A simple example: $f(x)=1$ and $g(x)=f(x)(1+sin(2piln x))$. Neither the limit at $0^+$ nor the limit at $infty$ exist.
answered Jan 24 at 15:51
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You should clarify the use of the squeeze theorem: since $0le 1+sin(2piln x)le 2$, we also have
$$
0le g(x)le 2f(x)
$$
and since $lim_{xto?}f(x)=0$ (where $?=infty$ or $?=0^+$), we can conclude that
$$
lim_{xto?}g(x)=0
$$
This uses substantially that the limits of $f$ are both zero, otherwise the application of the squeeze theorem cannot be done. A simple example: $f(x)=1$ and $g(x)=f(x)(1+sin(2piln x))$. Neither the limit at $0^+$ nor the limit at $infty$ exist.
answered Jan 24 at 15:51
egregegreg
184k1486205
$endgroup$
You should clarify the use of the squeeze theorem: since $0le 1+sin(2piln x)le 2$, we also have
$$
0le g(x)le 2f(x)
$$
and since $lim_{xto?}f(x)=0$ (where $?=infty$ or $?=0^+$), we can conclude that
$$
lim_{xto?}g(x)=0
$$
This uses substantially that the limits of $f$ are both zero, otherwise the application of the squeeze theorem cannot be done. A simple example: $f(x)=1$ and $g(x)=f(x)(1+sin(2piln x))$. Neither the limit at $0^+$ nor the limit at $infty$ exist.
answered Jan 24 at 15:51
egregegreg
184k1486205
answered Jan 24 at 15:51
egregegreg
184k1486205
answered Jan 24 at 15:51
egregegreg
184k1486205
answered Jan 24 at 15:51
egregegreg
184k1486205
184k1486205
add a comment |
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$begingroup$
This is correct, although I would highlight that $g$ is the product of $f$ by something that is bounded.
$endgroup$
– Fimpellizieri
May 22 '18 at 1:29
1
$begingroup$
You can make more explicit what you are using. For example, the inequalities that you wrote imply $0leq |g(x)|=|f(x)||1+sin(2piln(x))|leq 2|f(x)|$. Then, by the whichever-way-you-call-it theorem, the limit of $|g(x)|$ is zero. This implies that the limit of $g(x)$ is also zero.
$endgroup$
– user561348
May 22 '18 at 1:29
$begingroup$
$g(x) = f(x)h(x).$ If $lim_limits{xto a} f(x), lim_limits{xto a} h(x)$ exist $lim_limits{xto a} g(x) = (lim_limits{xto a} f(x) )(lim_limits{xto a} h(x)), h(x)$ is bounded. i.e. $0le h(x)le 2, 0<lim_limits{xto a} g(x) <2lim_limits{xto a} f(x)$
$endgroup$
– Doug M
May 22 '18 at 1:36