Mixing
Multi variable Limit to $infty$
$begingroup$
The question says that if $frac{cos x}{sin ax}$ is a periodic function then find the find the value of $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi aright)$$
I really don't have any clue on how to begin the question. I have never seen a question like this. The only thing I could think of since is that since $frac{cos x}{sin ax}$ is periodic then we can assume $a$ to be $1$ since $cot x$ is periodic. So then we have to find the limit of the function, $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi right)$$ But even this is based on an assumption and I have no idea on how to go from here. Any clue or hint would be much appreciated.
calculus limits
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
The question says that if $frac{cos x}{sin ax}$ is a periodic function then find the find the value of $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi aright)$$
I really don't have any clue on how to begin the question. I have never seen a question like this. The only thing I could think of since is that since $frac{cos x}{sin ax}$ is periodic then we can assume $a$ to be $1$ since $cot x$ is periodic. So then we have to find the limit of the function, $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi right)$$ But even this is based on an assumption and I have no idea on how to go from here. Any clue or hint would be much appreciated.
calculus limits
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
The question says that if $frac{cos x}{sin ax}$ is a periodic function then find the find the value of $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi aright)$$
I really don't have any clue on how to begin the question. I have never seen a question like this. The only thing I could think of since is that since $frac{cos x}{sin ax}$ is periodic then we can assume $a$ to be $1$ since $cot x$ is periodic. So then we have to find the limit of the function, $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi right)$$ But even this is based on an assumption and I have no idea on how to go from here. Any clue or hint would be much appreciated.
calculus limits
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
The question says that if $frac{cos x}{sin ax}$ is a periodic function then find the find the value of $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi aright)$$
I really don't have any clue on how to begin the question. I have never seen a question like this. The only thing I could think of since is that since $frac{cos x}{sin ax}$ is periodic then we can assume $a$ to be $1$ since $cot x$ is periodic. So then we have to find the limit of the function, $$lim _{m rightarrow infty} lim_{n rightarrow infty} left(1+ cos^{2m} n! ; pi right)$$ But even this is based on an assumption and I have no idea on how to go from here. Any clue or hint would be much appreciated.
calculus limits
calculus limits
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
asked Jan 28 at 14:26
Prakhar NagpalPrakhar Nagpal
752318
752318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
... we can assume $a$ to be 1...
We can (must) assume $a$ to be rational.
In your case (the general case is almost the same) the inner limit
$$lim_{nrightarrow infty}left(1+cos^{2m}n!piright)$$
is trivial because for $kinBbb Z$, $cos(kpi) = cdots$
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
$begingroup$
Alright so we can just say that since anything of the form $cos kx = 1$ then we can say the limit of the function is $1+1=2$?
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
I did actually think of the problem in that way but I just wasn't convinced by that argument
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
@PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:41
$begingroup$
Alright, thank you so much!
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:42
$begingroup$
@PrakharNagpal, $cos kpi = pm 1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:42
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
... we can assume $a$ to be 1...
We can (must) assume $a$ to be rational.
In your case (the general case is almost the same) the inner limit
$$lim_{nrightarrow infty}left(1+cos^{2m}n!piright)$$
is trivial because for $kinBbb Z$, $cos(kpi) = cdots$
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
$begingroup$
Alright so we can just say that since anything of the form $cos kx = 1$ then we can say the limit of the function is $1+1=2$?
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
I did actually think of the problem in that way but I just wasn't convinced by that argument
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
@PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:41
$begingroup$
Alright, thank you so much!
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:42
$begingroup$
@PrakharNagpal, $cos kpi = pm 1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:42
add a comment |
$begingroup$
... we can assume $a$ to be 1...
We can (must) assume $a$ to be rational.
In your case (the general case is almost the same) the inner limit
$$lim_{nrightarrow infty}left(1+cos^{2m}n!piright)$$
is trivial because for $kinBbb Z$, $cos(kpi) = cdots$
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
$begingroup$
Alright so we can just say that since anything of the form $cos kx = 1$ then we can say the limit of the function is $1+1=2$?
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
I did actually think of the problem in that way but I just wasn't convinced by that argument
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
@PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:41
$begingroup$
Alright, thank you so much!
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:42
$begingroup$
@PrakharNagpal, $cos kpi = pm 1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:42
add a comment |
$begingroup$
... we can assume $a$ to be 1...
We can (must) assume $a$ to be rational.
In your case (the general case is almost the same) the inner limit
$$lim_{nrightarrow infty}left(1+cos^{2m}n!piright)$$
is trivial because for $kinBbb Z$, $cos(kpi) = cdots$
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
... we can assume $a$ to be 1...
We can (must) assume $a$ to be rational.
In your case (the general case is almost the same) the inner limit
$$lim_{nrightarrow infty}left(1+cos^{2m}n!piright)$$
is trivial because for $kinBbb Z$, $cos(kpi) = cdots$
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Jan 28 at 14:33
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
34.8k42971
$begingroup$
Alright so we can just say that since anything of the form $cos kx = 1$ then we can say the limit of the function is $1+1=2$?
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
I did actually think of the problem in that way but I just wasn't convinced by that argument
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
@PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:41
$begingroup$
Alright, thank you so much!
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:42
$begingroup$
@PrakharNagpal, $cos kpi = pm 1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:42
add a comment |
$begingroup$
Alright so we can just say that since anything of the form $cos kx = 1$ then we can say the limit of the function is $1+1=2$?
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
I did actually think of the problem in that way but I just wasn't convinced by that argument
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
@PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:41
$begingroup$
Alright, thank you so much!
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:42
$begingroup$
@PrakharNagpal, $cos kpi = pm 1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:42
$begingroup$
Alright so we can just say that since anything of the form $cos kx = 1$ then we can say the limit of the function is $1+1=2$?
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
Alright so we can just say that since anything of the form $cos kx = 1$ then we can say the limit of the function is $1+1=2$?
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
I did actually think of the problem in that way but I just wasn't convinced by that argument
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
I did actually think of the problem in that way but I just wasn't convinced by that argument
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:39
$begingroup$
@PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:41
$begingroup$
@PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:41
$begingroup$
Alright, thank you so much!
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:42
$begingroup$
Alright, thank you so much!
$endgroup$
– Prakhar Nagpal
Jan 28 at 14:42
$begingroup$
@PrakharNagpal, $cos kpi = pm 1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:42
$begingroup$
@PrakharNagpal, $cos kpi = pm 1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 28 at 14:42
add a comment |
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