Compute the limit or show it doesn't exist: $lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n})$
$begingroup$
Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$
I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.
calculus limits trigonometry
edited Dec 18 '18 at 4:24
tatan
5,63362655
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
$endgroup$
add a comment |
$begingroup$
Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$
I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.
calculus limits trigonometry
edited Dec 18 '18 at 4:24
tatan
5,63362655
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
$endgroup$
2
$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05
1
$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10
add a comment |
$begingroup$
Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$
I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.
calculus limits trigonometry
edited Dec 18 '18 at 4:24
tatan
5,63362655
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
$endgroup$
Determine the following limit, or show it doesn't exist:
$$lim_{nto infty}(sinsqrt{n+1} - sinsqrt{n}) .$$
I'm not sure how to proceed. I know that I can't use limit arithmetic because both $lim_{nto infty}sinsqrt{n+1}$ and $lim_{nto infty}sinsqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.
calculus limits trigonometry
calculus limits trigonometry
edited Dec 18 '18 at 4:24
tatan
5,63362655
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
edited Dec 18 '18 at 4:24
tatan
5,63362655
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
edited Dec 18 '18 at 4:24
tatan
5,63362655
edited Dec 18 '18 at 4:24
tatan
5,63362655
edited Dec 18 '18 at 4:24
tatan
5,63362655
5,63362655
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
asked Dec 18 '18 at 4:02
J. LastinJ. Lastin
986
986
2
$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05
1
$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10
add a comment |
2
$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05
1
$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10
2
2
$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05
$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05
1
1
$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10
$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
$$|sin x - sin y| leq |x - y| .$$
Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
$endgroup$
add a comment |
$begingroup$
$sin x < x $
$|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
$endgroup$
add a comment |
$begingroup$
$$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$
For real $y,-1lecos yle1$
For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
By the mean value theorem,
$|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
$$|sin x - sin y| leq |x - y| .$$
Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
$endgroup$
add a comment |
$begingroup$
Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
$$|sin x - sin y| leq |x - y| .$$
Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
$endgroup$
add a comment |
$begingroup$
Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
$$|sin x - sin y| leq |x - y| .$$
Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
$endgroup$
Hint Since $sin$ is differentiable and $| {sin x} | leq 1$ for all (real) $x$, we have
$$|sin x - sin y| leq |x - y| .$$
Taking $x = sqrt{n + 1}$ and $y = sqrt{n}$ gives gives that the quantity $sin sqrt{n + 1} - sin sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$sqrt{n + 1} - sqrt{n} = frac{1}{sqrt{n + 1} + sqrt{n}} leq frac{1}{2 sqrt{n}} .$$
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
answered Dec 18 '18 at 4:18
TravisTravis
60k767146
60k767146
add a comment |
add a comment |
$begingroup$
$sin x < x $
$|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
$endgroup$
add a comment |
$begingroup$
$sin x < x $
$|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
$endgroup$
add a comment |
$begingroup$
$sin x < x $
$|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
$endgroup$
$sin x < x $
$|sin sqrt{x+1} - sin sqrt{x}| leq |sqrt{x+1} - sqrt{x}| = dfrac{1}{|sqrt{x+1} + sqrt{x}|} to 0$
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
answered Dec 18 '18 at 4:16
l''''''''ll''''''''l
2,198726
2,198726
add a comment |
add a comment |
$begingroup$
$$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$
For real $y,-1lecos yle1$
For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$
For real $y,-1lecos yle1$
For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$
For real $y,-1lecos yle1$
For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$$sinsqrt{x+1}-sinsqrt x=2sindfrac{sqrt{x+1}-sqrt x}2cosdfrac{sqrt{x+1}+sqrt x}2$$
For real $y,-1lecos yle1$
For $lim_{xtoinfty}sin(...)=lim_{...}sindfrac1{2(sqrt{x+1}+sqrt x)}=?$
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 18 '18 at 4:43
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$begingroup$
By the mean value theorem,
$|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
$endgroup$
add a comment |
$begingroup$
By the mean value theorem,
$|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
$endgroup$
add a comment |
$begingroup$
By the mean value theorem,
$|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
$endgroup$
By the mean value theorem,
$|sin(sqrt{n+1}) - sin(sqrt{n})| leq |cos(theta_n)||sqrt{n+1}-sqrt{n}|$ and $|cos(theta_n)| leq 1$ for all n. Now its easy to finish.
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
answered Dec 18 '18 at 4:53
Mustafa SaidMustafa Said
2,9511913
2,9511913
add a comment |
add a comment |
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2
$begingroup$
use the mean value theorem
$endgroup$
– user124910
Dec 18 '18 at 4:05
1
$begingroup$
$sinalpha-sinbeta=ldots$
$endgroup$
– Artem
Dec 18 '18 at 4:10