Mixing
$lim_{nto +infty} |(2npi+1/n) sin(2npi+1/n)|$, $nin mathbb N$
-1
$begingroup$
$lim_{nto +infty} |(2npi+1/n) sin(2npi+1/n)|$, $nin mathbb N$. I tried L'Hospital which gives
$$ lim_{nto +infty}left| frac{sin(2npi+1/n)}{(2npi+1/n)^{-1}} right| = lim_{nto +infty}left| frac{cos(frac{2pi x^2+1}{x})(2pi x^2+1)^2}{x^2} right|.$$
Is the limit $+infty$? I am not sure because of the cosine term.
calculus proof-verification
asked Feb 1 at 22:53
user398843user398843
716316
$endgroup$
|
show 5 more comments
-1
$begingroup$
$lim_{nto +infty} |(2npi+1/n) sin(2npi+1/n)|$, $nin mathbb N$. I tried L'Hospital which gives
$$ lim_{nto +infty}left| frac{sin(2npi+1/n)}{(2npi+1/n)^{-1}} right| = lim_{nto +infty}left| frac{cos(frac{2pi x^2+1}{x})(2pi x^2+1)^2}{x^2} right|.$$
Is the limit $+infty$? I am not sure because of the cosine term.
calculus proof-verification
asked Feb 1 at 22:53
user398843user398843
716316
$endgroup$
1
$begingroup$
When can you use the rule of L'Hospital?
$endgroup$
– babemcnuggets
Feb 1 at 23:09
$begingroup$
@babemcnuggets $|(2npi+1/n) sin(2npi+1/n)|$ is an indeterminate product $0 * infty$, so I change it to $frac{0}{0}$ and use L'Hospital.
$endgroup$
– user398843
Feb 1 at 23:14
$begingroup$
I only remember using L'Hospital for $frac{infty}{infty}$ and $frac{0}{0}$, I might be wrong..
$endgroup$
– babemcnuggets
Feb 1 at 23:21
$begingroup$
@babemcnuggets en.wikipedia.org/wiki/…
$endgroup$
– user398843
Feb 1 at 23:23
1
$begingroup$
@user398843 Thanks for the clarifications. Also, as I stated, note you should not try to use l'Hôpital's rule in that case. You might get the correct answer, but it's not guaranteed. Regarding your question title & text change, note that an easier way to state $n$ is a positive integer is that $n in mathbb{N}$, i.e., is a natural number.
$endgroup$
– John Omielan
Feb 2 at 0:11
|
show 5 more comments
-1
-1
-1
1
$begingroup$
$lim_{nto +infty} |(2npi+1/n) sin(2npi+1/n)|$, $nin mathbb N$. I tried L'Hospital which gives
$$ lim_{nto +infty}left| frac{sin(2npi+1/n)}{(2npi+1/n)^{-1}} right| = lim_{nto +infty}left| frac{cos(frac{2pi x^2+1}{x})(2pi x^2+1)^2}{x^2} right|.$$
Is the limit $+infty$? I am not sure because of the cosine term.
calculus proof-verification
asked Feb 1 at 22:53
user398843user398843
716316
$endgroup$
$lim_{nto +infty} |(2npi+1/n) sin(2npi+1/n)|$, $nin mathbb N$. I tried L'Hospital which gives
$$ lim_{nto +infty}left| frac{sin(2npi+1/n)}{(2npi+1/n)^{-1}} right| = lim_{nto +infty}left| frac{cos(frac{2pi x^2+1}{x})(2pi x^2+1)^2}{x^2} right|.$$
Is the limit $+infty$? I am not sure because of the cosine term.
calculus proof-verification
calculus proof-verification
asked Feb 1 at 22:53
user398843user398843
716316
asked Feb 1 at 22:53
user398843user398843
716316
edited Feb 2 at 0:23
user398843
asked Feb 1 at 22:53
user398843user398843
716316
asked Feb 1 at 22:53
user398843user398843
716316
asked Feb 1 at 22:53
user398843user398843
716316
716316
1
$begingroup$
When can you use the rule of L'Hospital?
$endgroup$
– babemcnuggets
Feb 1 at 23:09
$begingroup$
@babemcnuggets $|(2npi+1/n) sin(2npi+1/n)|$ is an indeterminate product $0 * infty$, so I change it to $frac{0}{0}$ and use L'Hospital.
$endgroup$
– user398843
Feb 1 at 23:14
$begingroup$
I only remember using L'Hospital for $frac{infty}{infty}$ and $frac{0}{0}$, I might be wrong..
$endgroup$
– babemcnuggets
Feb 1 at 23:21
$begingroup$
@babemcnuggets en.wikipedia.org/wiki/…
$endgroup$
– user398843
Feb 1 at 23:23
1
$begingroup$
@user398843 Thanks for the clarifications. Also, as I stated, note you should not try to use l'Hôpital's rule in that case. You might get the correct answer, but it's not guaranteed. Regarding your question title & text change, note that an easier way to state $n$ is a positive integer is that $n in mathbb{N}$, i.e., is a natural number.
$endgroup$
– John Omielan
Feb 2 at 0:11
|
show 5 more comments
1
$begingroup$
When can you use the rule of L'Hospital?
$endgroup$
– babemcnuggets
Feb 1 at 23:09
$begingroup$
@babemcnuggets $|(2npi+1/n) sin(2npi+1/n)|$ is an indeterminate product $0 * infty$, so I change it to $frac{0}{0}$ and use L'Hospital.
$endgroup$
– user398843
Feb 1 at 23:14
$begingroup$
I only remember using L'Hospital for $frac{infty}{infty}$ and $frac{0}{0}$, I might be wrong..
$endgroup$
– babemcnuggets
Feb 1 at 23:21
$begingroup$
@babemcnuggets en.wikipedia.org/wiki/…
$endgroup$
– user398843
Feb 1 at 23:23
1
$begingroup$
@user398843 Thanks for the clarifications. Also, as I stated, note you should not try to use l'Hôpital's rule in that case. You might get the correct answer, but it's not guaranteed. Regarding your question title & text change, note that an easier way to state $n$ is a positive integer is that $n in mathbb{N}$, i.e., is a natural number.
$endgroup$
– John Omielan
Feb 2 at 0:11
1
1
$begingroup$
When can you use the rule of L'Hospital?
$endgroup$
– babemcnuggets
Feb 1 at 23:09
$begingroup$
When can you use the rule of L'Hospital?
$endgroup$
– babemcnuggets
Feb 1 at 23:09
$begingroup$
@babemcnuggets $|(2npi+1/n) sin(2npi+1/n)|$ is an indeterminate product $0 * infty$, so I change it to $frac{0}{0}$ and use L'Hospital.
$endgroup$
– user398843
Feb 1 at 23:14
$begingroup$
@babemcnuggets $|(2npi+1/n) sin(2npi+1/n)|$ is an indeterminate product $0 * infty$, so I change it to $frac{0}{0}$ and use L'Hospital.
$endgroup$
– user398843
Feb 1 at 23:14
$begingroup$
I only remember using L'Hospital for $frac{infty}{infty}$ and $frac{0}{0}$, I might be wrong..
$endgroup$
– babemcnuggets
Feb 1 at 23:21
$begingroup$
I only remember using L'Hospital for $frac{infty}{infty}$ and $frac{0}{0}$, I might be wrong..
$endgroup$
– babemcnuggets
Feb 1 at 23:21
$begingroup$
@babemcnuggets en.wikipedia.org/wiki/…
$endgroup$
– user398843
Feb 1 at 23:23
$begingroup$
@babemcnuggets en.wikipedia.org/wiki/…
$endgroup$
– user398843
Feb 1 at 23:23
1
1
$begingroup$
@user398843 Thanks for the clarifications. Also, as I stated, note you should not try to use l'Hôpital's rule in that case. You might get the correct answer, but it's not guaranteed. Regarding your question title & text change, note that an easier way to state $n$ is a positive integer is that $n in mathbb{N}$, i.e., is a natural number.
$endgroup$
– John Omielan
Feb 2 at 0:11
$begingroup$
@user398843 Thanks for the clarifications. Also, as I stated, note you should not try to use l'Hôpital's rule in that case. You might get the correct answer, but it's not guaranteed. Regarding your question title & text change, note that an easier way to state $n$ is a positive integer is that $n in mathbb{N}$, i.e., is a natural number.
$endgroup$
– John Omielan
Feb 2 at 0:11
|
show 5 more comments
1 Answer
1
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3
$begingroup$
Note that $sin$ has period $2pi$, so $sin(2npi+frac 1 n)=sin(frac 1 n)$. $lim (|2npi+frac 1 n)sin(frac 1 n)|=lim (|2npi+frac 1 n)(frac 1 n)|=2pi$. I have used the fact that $frac {sin, x } x to 1$ as $ x to 0$.
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
$endgroup$
$begingroup$
Can we really take the limit sign into the absolute value sign? If $a_n = (-1)^n$, then $lim_{nto infty} |(-1)^n| = 1$ but $|lim_{nto infty} (-1)^n|$ does not exist. I asked this because you use $lim_{xto 0} frac{sin x}{x}$.
$endgroup$
– user398843
Feb 1 at 23:45
$begingroup$
@user398843 When $lim a_n$ exists we can take the limit inside. In our case the limit exists.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:47
add a comment |
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1 Answer
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3
$begingroup$
Note that $sin$ has period $2pi$, so $sin(2npi+frac 1 n)=sin(frac 1 n)$. $lim (|2npi+frac 1 n)sin(frac 1 n)|=lim (|2npi+frac 1 n)(frac 1 n)|=2pi$. I have used the fact that $frac {sin, x } x to 1$ as $ x to 0$.
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
$endgroup$
$begingroup$
Can we really take the limit sign into the absolute value sign? If $a_n = (-1)^n$, then $lim_{nto infty} |(-1)^n| = 1$ but $|lim_{nto infty} (-1)^n|$ does not exist. I asked this because you use $lim_{xto 0} frac{sin x}{x}$.
$endgroup$
– user398843
Feb 1 at 23:45
$begingroup$
@user398843 When $lim a_n$ exists we can take the limit inside. In our case the limit exists.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:47
add a comment |
3
$begingroup$
Note that $sin$ has period $2pi$, so $sin(2npi+frac 1 n)=sin(frac 1 n)$. $lim (|2npi+frac 1 n)sin(frac 1 n)|=lim (|2npi+frac 1 n)(frac 1 n)|=2pi$. I have used the fact that $frac {sin, x } x to 1$ as $ x to 0$.
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
$endgroup$
$begingroup$
Can we really take the limit sign into the absolute value sign? If $a_n = (-1)^n$, then $lim_{nto infty} |(-1)^n| = 1$ but $|lim_{nto infty} (-1)^n|$ does not exist. I asked this because you use $lim_{xto 0} frac{sin x}{x}$.
$endgroup$
– user398843
Feb 1 at 23:45
$begingroup$
@user398843 When $lim a_n$ exists we can take the limit inside. In our case the limit exists.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:47
add a comment |
3
3
3
$begingroup$
Note that $sin$ has period $2pi$, so $sin(2npi+frac 1 n)=sin(frac 1 n)$. $lim (|2npi+frac 1 n)sin(frac 1 n)|=lim (|2npi+frac 1 n)(frac 1 n)|=2pi$. I have used the fact that $frac {sin, x } x to 1$ as $ x to 0$.
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
$endgroup$
Note that $sin$ has period $2pi$, so $sin(2npi+frac 1 n)=sin(frac 1 n)$. $lim (|2npi+frac 1 n)sin(frac 1 n)|=lim (|2npi+frac 1 n)(frac 1 n)|=2pi$. I have used the fact that $frac {sin, x } x to 1$ as $ x to 0$.
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
answered Feb 1 at 23:22
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
74.3k53270
$begingroup$
Can we really take the limit sign into the absolute value sign? If $a_n = (-1)^n$, then $lim_{nto infty} |(-1)^n| = 1$ but $|lim_{nto infty} (-1)^n|$ does not exist. I asked this because you use $lim_{xto 0} frac{sin x}{x}$.
$endgroup$
– user398843
Feb 1 at 23:45
$begingroup$
@user398843 When $lim a_n$ exists we can take the limit inside. In our case the limit exists.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:47
add a comment |
$begingroup$
Can we really take the limit sign into the absolute value sign? If $a_n = (-1)^n$, then $lim_{nto infty} |(-1)^n| = 1$ but $|lim_{nto infty} (-1)^n|$ does not exist. I asked this because you use $lim_{xto 0} frac{sin x}{x}$.
$endgroup$
– user398843
Feb 1 at 23:45
$begingroup$
@user398843 When $lim a_n$ exists we can take the limit inside. In our case the limit exists.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:47
$begingroup$
Can we really take the limit sign into the absolute value sign? If $a_n = (-1)^n$, then $lim_{nto infty} |(-1)^n| = 1$ but $|lim_{nto infty} (-1)^n|$ does not exist. I asked this because you use $lim_{xto 0} frac{sin x}{x}$.
$endgroup$
– user398843
Feb 1 at 23:45
$begingroup$
Can we really take the limit sign into the absolute value sign? If $a_n = (-1)^n$, then $lim_{nto infty} |(-1)^n| = 1$ but $|lim_{nto infty} (-1)^n|$ does not exist. I asked this because you use $lim_{xto 0} frac{sin x}{x}$.
$endgroup$
– user398843
Feb 1 at 23:45
$begingroup$
@user398843 When $lim a_n$ exists we can take the limit inside. In our case the limit exists.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:47
$begingroup$
@user398843 When $lim a_n$ exists we can take the limit inside. In our case the limit exists.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:47
add a comment |
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1
$begingroup$
When can you use the rule of L'Hospital?
$endgroup$
– babemcnuggets
Feb 1 at 23:09
$begingroup$
@babemcnuggets $|(2npi+1/n) sin(2npi+1/n)|$ is an indeterminate product $0 * infty$, so I change it to $frac{0}{0}$ and use L'Hospital.
$endgroup$
– user398843
Feb 1 at 23:14
$begingroup$
I only remember using L'Hospital for $frac{infty}{infty}$ and $frac{0}{0}$, I might be wrong..
$endgroup$
– babemcnuggets
Feb 1 at 23:21
$begingroup$
@babemcnuggets en.wikipedia.org/wiki/…
$endgroup$
– user398843
Feb 1 at 23:23
1
$begingroup$
@user398843 Thanks for the clarifications. Also, as I stated, note you should not try to use l'Hôpital's rule in that case. You might get the correct answer, but it's not guaranteed. Regarding your question title & text change, note that an easier way to state $n$ is a positive integer is that $n in mathbb{N}$, i.e., is a natural number.
$endgroup$
– John Omielan
Feb 2 at 0:11