Show that the limit is divergent












1














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




















  • 1




    What is your definition of divergent? (Not convergent or limit infinite?)
    – Martín Vacas Vignolo
    May 6 at 20:56










  • You are right, I wrote it not precisely. I corrected what do I mean
    – MathMen
    May 6 at 20:59
















1














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




















  • 1




    What is your definition of divergent? (Not convergent or limit infinite?)
    – Martín Vacas Vignolo
    May 6 at 20:56










  • You are right, I wrote it not precisely. I corrected what do I mean
    – MathMen
    May 6 at 20:59














1












1








1







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















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 at 2:32









Robert Howard

1,9161822




1,9161822










asked May 6 at 20:53









MathMen

1217




1217








  • 1




    What is your definition of divergent? (Not convergent or limit infinite?)
    – Martín Vacas Vignolo
    May 6 at 20:56










  • You are right, I wrote it not precisely. I corrected what do I mean
    – MathMen
    May 6 at 20:59














  • 1




    What is your definition of divergent? (Not convergent or limit infinite?)
    – Martín Vacas Vignolo
    May 6 at 20:56










  • You are right, I wrote it not precisely. I corrected what do I mean
    – MathMen
    May 6 at 20:59








1




1




What is your definition of divergent? (Not convergent or limit infinite?)
– Martín Vacas Vignolo
May 6 at 20:56




What is your definition of divergent? (Not convergent or limit infinite?)
– Martín Vacas Vignolo
May 6 at 20:56












You are right, I wrote it not precisely. I corrected what do I mean
– MathMen
May 6 at 20:59




You are right, I wrote it not precisely. I corrected what do I mean
– MathMen
May 6 at 20:59










2 Answers
2






active

oldest

votes


















0














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





















  • 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 $xrightarrow 0^{+}$?
    – MathMen
    May 6 at 21:14












  • If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
    – gimusi
    May 6 at 21:15










  • But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
    – MathMen
    May 6 at 21:19










  • @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).
    – gimusi
    May 6 at 21:21





















0














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





















  • You are right of course!
    – gimusi
    May 6 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














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





















  • 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 $xrightarrow 0^{+}$?
    – MathMen
    May 6 at 21:14












  • If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
    – gimusi
    May 6 at 21:15










  • But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
    – MathMen
    May 6 at 21:19










  • @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).
    – gimusi
    May 6 at 21:21


















0














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





















  • 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 $xrightarrow 0^{+}$?
    – MathMen
    May 6 at 21:14












  • If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
    – gimusi
    May 6 at 21:15










  • But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
    – MathMen
    May 6 at 21:19










  • @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).
    – gimusi
    May 6 at 21:21
















0












0








0






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












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 at 21:08









gimusi

1




1












  • 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 $xrightarrow 0^{+}$?
    – MathMen
    May 6 at 21:14












  • If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
    – gimusi
    May 6 at 21:15










  • But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
    – MathMen
    May 6 at 21:19










  • @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).
    – gimusi
    May 6 at 21:21




















  • 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 $xrightarrow 0^{+}$?
    – MathMen
    May 6 at 21:14












  • If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
    – gimusi
    May 6 at 21:15










  • But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
    – MathMen
    May 6 at 21:19










  • @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).
    – gimusi
    May 6 at 21:21


















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 $xrightarrow 0^{+}$?
– MathMen
May 6 at 21:14






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 $xrightarrow 0^{+}$?
– MathMen
May 6 at 21:14














If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
– gimusi
May 6 at 21:15




If $x^bto 0$ you can easily conclude for convergence by squeeze theorem.
– gimusi
May 6 at 21:15












But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
– MathMen
May 6 at 21:19




But if b<0, then $lim_{xrightarrow 0^{+}}x^b = + infty$
– MathMen
May 6 at 21:19












@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).
– gimusi
May 6 at 21:21






@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).
– gimusi
May 6 at 21:21













0














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





















  • You are right of course!
    – gimusi
    May 6 at 21:14
















0














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





















  • You are right of course!
    – gimusi
    May 6 at 21:14














0












0








0






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












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 at 21:10









Yves Daoust

124k671221




124k671221












  • You are right of course!
    – gimusi
    May 6 at 21:14


















  • You are right of course!
    – gimusi
    May 6 at 21:14
















You are right of course!
– gimusi
May 6 at 21:14




You are right of course!
– gimusi
May 6 at 21:14


















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