Show that the limit is divergent












1












$begingroup$


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:




  1. $x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.


  2. $x = e^{frac{2n+1}{c}pi}$, 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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


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:




  1. $x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.


  2. $x = e^{frac{2n+1}{c}pi}$, 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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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:




  1. $x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.


  2. $x = e^{frac{2n+1}{c}pi}$, 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







share|cite|improve this question











$endgroup$




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:




  1. $x = e^{frac{2n}{c}pi}$, then $lim_{x rightarrow infty} A cos(c log(x)) = A$.


  2. $x = e^{frac{2n+1}{c}pi}$, 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 2:32









Robert Howard

1,9801822




1,9801822










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














  • 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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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 $-infty$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















0












$begingroup$

The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are right of course!
    $endgroup$
    – gimusi
    May 6 '18 at 21:14











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $-infty$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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 $-infty$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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 $-infty$)






share|cite|improve this answer









$endgroup$



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 $-infty$)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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













0












$begingroup$

The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are right of course!
    $endgroup$
    – gimusi
    May 6 '18 at 21:14
















0












$begingroup$

The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are right of course!
    $endgroup$
    – gimusi
    May 6 '18 at 21:14














0












0








0





$begingroup$

The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.






share|cite|improve this answer









$endgroup$



The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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