Prove that $lim_{x to 0} frac{x - sin x}{x^3} = frac16$ using notable limits [duplicate]
This question already has an answer here:
Determine $lim_{x to 0}{frac{x-sin{x}}{x^3}}=frac{1}{6}$, without L'Hospital or Taylor
5 answers
Can we prove that $lim_{x to 0} frac{x - sin x}{x^3} = frac 16$ using just algebra, trigonometric theorems and notable limits $left( mathrm{i.e. }quad frac{sin x}{x} to 1 quad mathrm{and} quad frac{1 - cos x}{x^2} to frac 12 right) $ and no l'Hospital rule, no series expansion, no Taylor series?
limits alternative-proof
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
marked as duplicate by ncmathsadist, Community♦ Nov 22 '18 at 23:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Determine $lim_{x to 0}{frac{x-sin{x}}{x^3}}=frac{1}{6}$, without L'Hospital or Taylor
5 answers
Can we prove that $lim_{x to 0} frac{x - sin x}{x^3} = frac 16$ using just algebra, trigonometric theorems and notable limits $left( mathrm{i.e. }quad frac{sin x}{x} to 1 quad mathrm{and} quad frac{1 - cos x}{x^2} to frac 12 right) $ and no l'Hospital rule, no series expansion, no Taylor series?
limits alternative-proof
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
marked as duplicate by ncmathsadist, Community♦ Nov 22 '18 at 23:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
math.stackexchange.com/questions/217081/…
– aleden
Nov 22 '18 at 22:53
More in general refer to Are all limits solvable without L'Hôpital Rule or Series Expansion
– gimusi
Nov 22 '18 at 23:08
add a comment |
This question already has an answer here:
Determine $lim_{x to 0}{frac{x-sin{x}}{x^3}}=frac{1}{6}$, without L'Hospital or Taylor
5 answers
Can we prove that $lim_{x to 0} frac{x - sin x}{x^3} = frac 16$ using just algebra, trigonometric theorems and notable limits $left( mathrm{i.e. }quad frac{sin x}{x} to 1 quad mathrm{and} quad frac{1 - cos x}{x^2} to frac 12 right) $ and no l'Hospital rule, no series expansion, no Taylor series?
limits alternative-proof
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
This question already has an answer here:
Determine $lim_{x to 0}{frac{x-sin{x}}{x^3}}=frac{1}{6}$, without L'Hospital or Taylor
5 answers
Can we prove that $lim_{x to 0} frac{x - sin x}{x^3} = frac 16$ using just algebra, trigonometric theorems and notable limits $left( mathrm{i.e. }quad frac{sin x}{x} to 1 quad mathrm{and} quad frac{1 - cos x}{x^2} to frac 12 right) $ and no l'Hospital rule, no series expansion, no Taylor series?
This question already has an answer here:
Determine $lim_{x to 0}{frac{x-sin{x}}{x^3}}=frac{1}{6}$, without L'Hospital or Taylor
5 answers
limits alternative-proof
limits alternative-proof
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
asked Nov 22 '18 at 22:50
yamatteoyamatteo
1
1
marked as duplicate by ncmathsadist, Community♦ Nov 22 '18 at 23:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by ncmathsadist, Community♦ Nov 22 '18 at 23:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
math.stackexchange.com/questions/217081/…
– aleden
Nov 22 '18 at 22:53
More in general refer to Are all limits solvable without L'Hôpital Rule or Series Expansion
– gimusi
Nov 22 '18 at 23:08
add a comment |
2
math.stackexchange.com/questions/217081/…
– aleden
Nov 22 '18 at 22:53
More in general refer to Are all limits solvable without L'Hôpital Rule or Series Expansion
– gimusi
Nov 22 '18 at 23:08
2
2
math.stackexchange.com/questions/217081/…
– aleden
Nov 22 '18 at 22:53
math.stackexchange.com/questions/217081/…
– aleden
Nov 22 '18 at 22:53
More in general refer to Are all limits solvable without L'Hôpital Rule or Series Expansion
– gimusi
Nov 22 '18 at 23:08
More in general refer to Are all limits solvable without L'Hôpital Rule or Series Expansion
– gimusi
Nov 22 '18 at 23:08
add a comment |
1 Answer
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$sin x=sin (x/3)cos(2x/3)+cos(x/3)sin(2x/3)=sin (x/3)(1-2sin^2(2x/3))+2cos^2(x/3)sin(x/3)=sin (x/3)-2sin^3 (x/3)+2sin (x/3)-2sin^3 (x/3)=3sin (x/3)-4sin^3 (x/3)$
$frac{x-sin x}{x^3}=frac{x-3sin (x/3)-4sin^3 (x/3)}{x^3}=frac{x-3sin (x/3)}{x^3}-frac{4sin^3 (x/3)}{x^3}=1/9frac{x/3-sin (x/3)}{(x/3)^3}-4/27 frac{sin^3 (x/3)}{(x/3)^3}$
$m=lim_{xrightarrow 0} frac{x-sin x}{x^3}= 1/9lim_{xrightarrow 0}frac{x/3-sin (x/3)}{(x/3)^3}-4/27lim_{xrightarrow 0} frac{sin^3 (x/3)}{(x/3)^3}=1/9M-4/27 Rightarrow 8/9M=4/27Rightarrow M=1/6.$
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
This is a really nifty way to compute the limit, but it assumes the limit exists.
– Barry Cipra
Nov 22 '18 at 23:15
Yes, that's correct.
– John_Wick
Nov 22 '18 at 23:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$sin x=sin (x/3)cos(2x/3)+cos(x/3)sin(2x/3)=sin (x/3)(1-2sin^2(2x/3))+2cos^2(x/3)sin(x/3)=sin (x/3)-2sin^3 (x/3)+2sin (x/3)-2sin^3 (x/3)=3sin (x/3)-4sin^3 (x/3)$
$frac{x-sin x}{x^3}=frac{x-3sin (x/3)-4sin^3 (x/3)}{x^3}=frac{x-3sin (x/3)}{x^3}-frac{4sin^3 (x/3)}{x^3}=1/9frac{x/3-sin (x/3)}{(x/3)^3}-4/27 frac{sin^3 (x/3)}{(x/3)^3}$
$m=lim_{xrightarrow 0} frac{x-sin x}{x^3}= 1/9lim_{xrightarrow 0}frac{x/3-sin (x/3)}{(x/3)^3}-4/27lim_{xrightarrow 0} frac{sin^3 (x/3)}{(x/3)^3}=1/9M-4/27 Rightarrow 8/9M=4/27Rightarrow M=1/6.$
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
This is a really nifty way to compute the limit, but it assumes the limit exists.
– Barry Cipra
Nov 22 '18 at 23:15
Yes, that's correct.
– John_Wick
Nov 22 '18 at 23:16
add a comment |
$sin x=sin (x/3)cos(2x/3)+cos(x/3)sin(2x/3)=sin (x/3)(1-2sin^2(2x/3))+2cos^2(x/3)sin(x/3)=sin (x/3)-2sin^3 (x/3)+2sin (x/3)-2sin^3 (x/3)=3sin (x/3)-4sin^3 (x/3)$
$frac{x-sin x}{x^3}=frac{x-3sin (x/3)-4sin^3 (x/3)}{x^3}=frac{x-3sin (x/3)}{x^3}-frac{4sin^3 (x/3)}{x^3}=1/9frac{x/3-sin (x/3)}{(x/3)^3}-4/27 frac{sin^3 (x/3)}{(x/3)^3}$
$m=lim_{xrightarrow 0} frac{x-sin x}{x^3}= 1/9lim_{xrightarrow 0}frac{x/3-sin (x/3)}{(x/3)^3}-4/27lim_{xrightarrow 0} frac{sin^3 (x/3)}{(x/3)^3}=1/9M-4/27 Rightarrow 8/9M=4/27Rightarrow M=1/6.$
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
This is a really nifty way to compute the limit, but it assumes the limit exists.
– Barry Cipra
Nov 22 '18 at 23:15
Yes, that's correct.
– John_Wick
Nov 22 '18 at 23:16
add a comment |
$sin x=sin (x/3)cos(2x/3)+cos(x/3)sin(2x/3)=sin (x/3)(1-2sin^2(2x/3))+2cos^2(x/3)sin(x/3)=sin (x/3)-2sin^3 (x/3)+2sin (x/3)-2sin^3 (x/3)=3sin (x/3)-4sin^3 (x/3)$
$frac{x-sin x}{x^3}=frac{x-3sin (x/3)-4sin^3 (x/3)}{x^3}=frac{x-3sin (x/3)}{x^3}-frac{4sin^3 (x/3)}{x^3}=1/9frac{x/3-sin (x/3)}{(x/3)^3}-4/27 frac{sin^3 (x/3)}{(x/3)^3}$
$m=lim_{xrightarrow 0} frac{x-sin x}{x^3}= 1/9lim_{xrightarrow 0}frac{x/3-sin (x/3)}{(x/3)^3}-4/27lim_{xrightarrow 0} frac{sin^3 (x/3)}{(x/3)^3}=1/9M-4/27 Rightarrow 8/9M=4/27Rightarrow M=1/6.$
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
$sin x=sin (x/3)cos(2x/3)+cos(x/3)sin(2x/3)=sin (x/3)(1-2sin^2(2x/3))+2cos^2(x/3)sin(x/3)=sin (x/3)-2sin^3 (x/3)+2sin (x/3)-2sin^3 (x/3)=3sin (x/3)-4sin^3 (x/3)$
$frac{x-sin x}{x^3}=frac{x-3sin (x/3)-4sin^3 (x/3)}{x^3}=frac{x-3sin (x/3)}{x^3}-frac{4sin^3 (x/3)}{x^3}=1/9frac{x/3-sin (x/3)}{(x/3)^3}-4/27 frac{sin^3 (x/3)}{(x/3)^3}$
$m=lim_{xrightarrow 0} frac{x-sin x}{x^3}= 1/9lim_{xrightarrow 0}frac{x/3-sin (x/3)}{(x/3)^3}-4/27lim_{xrightarrow 0} frac{sin^3 (x/3)}{(x/3)^3}=1/9M-4/27 Rightarrow 8/9M=4/27Rightarrow M=1/6.$
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
answered Nov 22 '18 at 23:09
John_WickJohn_Wick
1,476111
1,476111
This is a really nifty way to compute the limit, but it assumes the limit exists.
– Barry Cipra
Nov 22 '18 at 23:15
Yes, that's correct.
– John_Wick
Nov 22 '18 at 23:16
add a comment |
This is a really nifty way to compute the limit, but it assumes the limit exists.
– Barry Cipra
Nov 22 '18 at 23:15
Yes, that's correct.
– John_Wick
Nov 22 '18 at 23:16
This is a really nifty way to compute the limit, but it assumes the limit exists.
– Barry Cipra
Nov 22 '18 at 23:15
This is a really nifty way to compute the limit, but it assumes the limit exists.
– Barry Cipra
Nov 22 '18 at 23:15
Yes, that's correct.
– John_Wick
Nov 22 '18 at 23:16
Yes, that's correct.
– John_Wick
Nov 22 '18 at 23:16
add a comment |
2
math.stackexchange.com/questions/217081/…
– aleden
Nov 22 '18 at 22:53
More in general refer to Are all limits solvable without L'Hôpital Rule or Series Expansion
– gimusi
Nov 22 '18 at 23:08