A solution for a limit
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What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
61
$endgroup$
add a comment |
$begingroup$
What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
61
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1
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You did not prove that there was a limit. It is fine apart from that.
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– Mindlack
Jan 9 at 14:43
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How can you write $sin 6y=3sin 2y$?
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– pie314271
Jan 9 at 14:44
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@pie314271 He's using the triple angle formula for $sin$
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– saulspatz
Jan 9 at 14:45
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(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
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– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
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– lab bhattacharjee
Jan 9 at 15:06
add a comment |
$begingroup$
What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
61
$endgroup$
What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
61
asked Jan 9 at 14:40
G. VasiG. Vasi
61
asked Jan 9 at 14:40
G. VasiG. Vasi
61
asked Jan 9 at 14:40
G. VasiG. Vasi
61
asked Jan 9 at 14:40
G. VasiG. Vasi
61
61
1
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You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
add a comment |
1
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
1
1
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
add a comment |
1 Answer
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$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
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add a comment |
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$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
$endgroup$
add a comment |
$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
$endgroup$
add a comment |
$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
$endgroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1794
1794
add a comment |
add a comment |
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1
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06