Show that the limit is divergent
$begingroup$
I want to calculate the following limit:
$$
lim_{x rightarrow infty} Ax^{b} cos(c log(x)),
$$
where $A$, $b>0$ and $c$ are some constants.
I suppose that the function $Ax^{b} cos(c log(x))$ is not convergent when $xrightarrow infty$.
Firstly, I think I can show that
$$
Acos(c log(x))
$$
is not convergent when $xrightarrow infty$.
For this purpose I use two different subsequences and show that the limit converges to different values:
$x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.
$x = e^{frac{2n+1}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = -A$.
Of course, I assumed that $Aneq 0$.
Moreover, I know that:
$$
lim_{x rightarrow infty} x^{b} = +infty
$$
for $b>0$
Then, I claim that:
$$
Ax^{b} cos(c log(x))
$$
is not convergent when $xrightarrow infty$, because it is a product of a function which is not convergent and a function which has a limit equal to $infty$.
Is this solution correct?
limits
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
asked May 6 '18 at 20:53
MathMenMathMen
1267
$endgroup$
add a comment |
$begingroup$
I want to calculate the following limit:
$$
lim_{x rightarrow infty} Ax^{b} cos(c log(x)),
$$
where $A$, $b>0$ and $c$ are some constants.
I suppose that the function $Ax^{b} cos(c log(x))$ is not convergent when $xrightarrow infty$.
Firstly, I think I can show that
$$
Acos(c log(x))
$$
is not convergent when $xrightarrow infty$.
For this purpose I use two different subsequences and show that the limit converges to different values:
$x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.
$x = e^{frac{2n+1}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = -A$.
Of course, I assumed that $Aneq 0$.
Moreover, I know that:
$$
lim_{x rightarrow infty} x^{b} = +infty
$$
for $b>0$
Then, I claim that:
$$
Ax^{b} cos(c log(x))
$$
is not convergent when $xrightarrow infty$, because it is a product of a function which is not convergent and a function which has a limit equal to $infty$.
Is this solution correct?
limits
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
asked May 6 '18 at 20:53
MathMenMathMen
1267
$endgroup$
1
$begingroup$
What is your definition of divergent? (Not convergent or limit infinite?)
$endgroup$
– Martín Vacas Vignolo
May 6 '18 at 20:56
$begingroup$
You are right, I wrote it not precisely. I corrected what do I mean
$endgroup$
– MathMen
May 6 '18 at 20:59
add a comment |
$begingroup$
I want to calculate the following limit:
$$
lim_{x rightarrow infty} Ax^{b} cos(c log(x)),
$$
where $A$, $b>0$ and $c$ are some constants.
I suppose that the function $Ax^{b} cos(c log(x))$ is not convergent when $xrightarrow infty$.
Firstly, I think I can show that
$$
Acos(c log(x))
$$
is not convergent when $xrightarrow infty$.
For this purpose I use two different subsequences and show that the limit converges to different values:
$x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.
$x = e^{frac{2n+1}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = -A$.
Of course, I assumed that $Aneq 0$.
Moreover, I know that:
$$
lim_{x rightarrow infty} x^{b} = +infty
$$
for $b>0$
Then, I claim that:
$$
Ax^{b} cos(c log(x))
$$
is not convergent when $xrightarrow infty$, because it is a product of a function which is not convergent and a function which has a limit equal to $infty$.
Is this solution correct?
limits
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
asked May 6 '18 at 20:53
MathMenMathMen
1267
$endgroup$
I want to calculate the following limit:
$$
lim_{x rightarrow infty} Ax^{b} cos(c log(x)),
$$
where $A$, $b>0$ and $c$ are some constants.
I suppose that the function $Ax^{b} cos(c log(x))$ is not convergent when $xrightarrow infty$.
Firstly, I think I can show that
$$
Acos(c log(x))
$$
is not convergent when $xrightarrow infty$.
For this purpose I use two different subsequences and show that the limit converges to different values:
$x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.
$x = e^{frac{2n+1}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = -A$.
Of course, I assumed that $Aneq 0$.
Moreover, I know that:
$$
lim_{x rightarrow infty} x^{b} = +infty
$$
for $b>0$
Then, I claim that:
$$
Ax^{b} cos(c log(x))
$$
is not convergent when $xrightarrow infty$, because it is a product of a function which is not convergent and a function which has a limit equal to $infty$.
Is this solution correct?
limits
limits
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
asked May 6 '18 at 20:53
MathMenMathMen
1267
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
asked May 6 '18 at 20:53
MathMenMathMen
1267
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
edited Dec 2 '18 at 2:32
Robert Howard
1,9801822
1,9801822
asked May 6 '18 at 20:53
MathMenMathMen
1267
asked May 6 '18 at 20:53
MathMenMathMen
1267
asked May 6 '18 at 20:53
MathMenMathMen
1267
1267
1
$begingroup$
What is your definition of divergent? (Not convergent or limit infinite?)
$endgroup$
– Martín Vacas Vignolo
May 6 '18 at 20:56
$begingroup$
You are right, I wrote it not precisely. I corrected what do I mean
$endgroup$
– MathMen
May 6 '18 at 20:59
add a comment |
1
$begingroup$
What is your definition of divergent? (Not convergent or limit infinite?)
$endgroup$
– Martín Vacas Vignolo
May 6 '18 at 20:56
$begingroup$
You are right, I wrote it not precisely. I corrected what do I mean
$endgroup$
– MathMen
May 6 '18 at 20:59
1
1
$begingroup$
What is your definition of divergent? (Not convergent or limit infinite?)
$endgroup$
– Martín Vacas Vignolo
May 6 '18 at 20:56
$begingroup$
What is your definition of divergent? (Not convergent or limit infinite?)
$endgroup$
– Martín Vacas Vignolo
May 6 '18 at 20:56
$begingroup$
You are right, I wrote it not precisely. I corrected what do I mean
$endgroup$
– MathMen
May 6 '18 at 20:59
$begingroup$
You are right, I wrote it not precisely. I corrected what do I mean
$endgroup$
– MathMen
May 6 '18 at 20:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes it is correct, we can simply note that since $|c log x|to infty$ is continuos
- $cos(c log(x))$ oscillates from $-1$ and $1$
and
- $|Ax^{b}|to infty$
then
- $Ax^{b} cos(c log(x))$ doesn't converges ("oscillates between" $+infty$ and $-infty$)
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Thank you! If I will assume now that $b<0$ and consider convergence: $xrightarrow 0^{+}$ can I draw the same conclusion, that is: that the function $Ax^b cos(clog(b)$ does not converges for $b<0$ when $xrightarrow 0^{+}$?
$endgroup$
– MathMen
May 6 '18 at 21:14
$begingroup$
If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
$endgroup$
– gimusi
May 6 '18 at 21:15
$begingroup$
But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
$endgroup$
– MathMen
May 6 '18 at 21:19
$begingroup$
@JonasAl-Hadad Ops sorry I was thinking to the case $to infty$! Yes of course in this case the same result holds since $cos$ oscillates and the other part diverges (or doesn't tend to zero).
$endgroup$
– gimusi
May 6 '18 at 21:21
add a comment |
$begingroup$
The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
$endgroup$
$begingroup$
You are right of course!
$endgroup$
– gimusi
May 6 '18 at 21:14
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes it is correct, we can simply note that since $|c log x|to infty$ is continuos
- $cos(c log(x))$ oscillates from $-1$ and $1$
and
- $|Ax^{b}|to infty$
then
- $Ax^{b} cos(c log(x))$ doesn't converges ("oscillates between" $+infty$ and $-infty$)
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Thank you! If I will assume now that $b<0$ and consider convergence: $xrightarrow 0^{+}$ can I draw the same conclusion, that is: that the function $Ax^b cos(clog(b)$ does not converges for $b<0$ when $xrightarrow 0^{+}$?
$endgroup$
– MathMen
May 6 '18 at 21:14
$begingroup$
If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
$endgroup$
– gimusi
May 6 '18 at 21:15
$begingroup$
But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
$endgroup$
– MathMen
May 6 '18 at 21:19
$begingroup$
@JonasAl-Hadad Ops sorry I was thinking to the case $to infty$! Yes of course in this case the same result holds since $cos$ oscillates and the other part diverges (or doesn't tend to zero).
$endgroup$
– gimusi
May 6 '18 at 21:21
add a comment |
$begingroup$
Yes it is correct, we can simply note that since $|c log x|to infty$ is continuos
- $cos(c log(x))$ oscillates from $-1$ and $1$
and
- $|Ax^{b}|to infty$
then
- $Ax^{b} cos(c log(x))$ doesn't converges ("oscillates between" $+infty$ and $-infty$)
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Thank you! If I will assume now that $b<0$ and consider convergence: $xrightarrow 0^{+}$ can I draw the same conclusion, that is: that the function $Ax^b cos(clog(b)$ does not converges for $b<0$ when $xrightarrow 0^{+}$?
$endgroup$
– MathMen
May 6 '18 at 21:14
$begingroup$
If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
$endgroup$
– gimusi
May 6 '18 at 21:15
$begingroup$
But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
$endgroup$
– MathMen
May 6 '18 at 21:19
$begingroup$
@JonasAl-Hadad Ops sorry I was thinking to the case $to infty$! Yes of course in this case the same result holds since $cos$ oscillates and the other part diverges (or doesn't tend to zero).
$endgroup$
– gimusi
May 6 '18 at 21:21
add a comment |
$begingroup$
Yes it is correct, we can simply note that since $|c log x|to infty$ is continuos
- $cos(c log(x))$ oscillates from $-1$ and $1$
and
- $|Ax^{b}|to infty$
then
- $Ax^{b} cos(c log(x))$ doesn't converges ("oscillates between" $+infty$ and $-infty$)
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
$endgroup$
Yes it is correct, we can simply note that since $|c log x|to infty$ is continuos
- $cos(c log(x))$ oscillates from $-1$ and $1$
and
- $|Ax^{b}|to infty$
then
- $Ax^{b} cos(c log(x))$ doesn't converges ("oscillates between" $+infty$ and $-infty$)
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
answered May 6 '18 at 21:08
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
Thank you! If I will assume now that $b<0$ and consider convergence: $xrightarrow 0^{+}$ can I draw the same conclusion, that is: that the function $Ax^b cos(clog(b)$ does not converges for $b<0$ when $xrightarrow 0^{+}$?
$endgroup$
– MathMen
May 6 '18 at 21:14
$begingroup$
If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
$endgroup$
– gimusi
May 6 '18 at 21:15
$begingroup$
But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
$endgroup$
– MathMen
May 6 '18 at 21:19
$begingroup$
@JonasAl-Hadad Ops sorry I was thinking to the case $to infty$! Yes of course in this case the same result holds since $cos$ oscillates and the other part diverges (or doesn't tend to zero).
$endgroup$
– gimusi
May 6 '18 at 21:21
add a comment |
$begingroup$
Thank you! If I will assume now that $b<0$ and consider convergence: $xrightarrow 0^{+}$ can I draw the same conclusion, that is: that the function $Ax^b cos(clog(b)$ does not converges for $b<0$ when $xrightarrow 0^{+}$?
$endgroup$
– MathMen
May 6 '18 at 21:14
$begingroup$
If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
$endgroup$
– gimusi
May 6 '18 at 21:15
$begingroup$
But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
$endgroup$
– MathMen
May 6 '18 at 21:19
$begingroup$
@JonasAl-Hadad Ops sorry I was thinking to the case $to infty$! Yes of course in this case the same result holds since $cos$ oscillates and the other part diverges (or doesn't tend to zero).
$endgroup$
– gimusi
May 6 '18 at 21:21
$begingroup$
Thank you! If I will assume now that $b<0$ and consider convergence: $xrightarrow 0^{+}$ can I draw the same conclusion, that is: that the function $Ax^b cos(clog(b)$ does not converges for $b<0$ when $xrightarrow 0^{+}$?
$endgroup$
– MathMen
May 6 '18 at 21:14
$begingroup$
Thank you! If I will assume now that $b<0$ and consider convergence: $xrightarrow 0^{+}$ can I draw the same conclusion, that is: that the function $Ax^b cos(clog(b)$ does not converges for $b<0$ when $xrightarrow 0^{+}$?
$endgroup$
– MathMen
May 6 '18 at 21:14
$begingroup$
If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
$endgroup$
– gimusi
May 6 '18 at 21:15
$begingroup$
If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
$endgroup$
– gimusi
May 6 '18 at 21:15
$begingroup$
But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
$endgroup$
– MathMen
May 6 '18 at 21:19
$begingroup$
But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
$endgroup$
– MathMen
May 6 '18 at 21:19
$begingroup$
@JonasAl-Hadad Ops sorry I was thinking to the case $to infty$! Yes of course in this case the same result holds since $cos$ oscillates and the other part diverges (or doesn't tend to zero).
$endgroup$
– gimusi
May 6 '18 at 21:21
$begingroup$
@JonasAl-Hadad Ops sorry I was thinking to the case $to infty$! Yes of course in this case the same result holds since $cos$ oscillates and the other part diverges (or doesn't tend to zero).
$endgroup$
– gimusi
May 6 '18 at 21:21
add a comment |
$begingroup$
The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
$endgroup$
$begingroup$
You are right of course!
$endgroup$
– gimusi
May 6 '18 at 21:14
add a comment |
$begingroup$
The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
$endgroup$
$begingroup$
You are right of course!
$endgroup$
– gimusi
May 6 '18 at 21:14
add a comment |
$begingroup$
The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
$endgroup$
The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
answered May 6 '18 at 21:10
Yves DaoustYves Daoust
128k674226
128k674226
$begingroup$
You are right of course!
$endgroup$
– gimusi
May 6 '18 at 21:14
add a comment |
$begingroup$
You are right of course!
$endgroup$
– gimusi
May 6 '18 at 21:14
$begingroup$
You are right of course!
$endgroup$
– gimusi
May 6 '18 at 21:14
$begingroup$
You are right of course!
$endgroup$
– gimusi
May 6 '18 at 21:14
add a comment |
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1
$begingroup$
What is your definition of divergent? (Not convergent or limit infinite?)
$endgroup$
– Martín Vacas Vignolo
May 6 '18 at 20:56
$begingroup$
You are right, I wrote it not precisely. I corrected what do I mean
$endgroup$
– MathMen
May 6 '18 at 20:59