Mixing
Why does $limlimits_{xrightarrow0}frac{1}{2x-1} log(2^{1+sin{x}}-1) = 2 $?
$begingroup$
Why does $limlimits_{xrightarrow0}frac{1}{2x-1} log(2^{1+sin{x}}-1)= 2 $?
I have:
$lim_{xrightarrow0}frac{1}{2x-1} = -1 $
$lim_{xrightarrow0}2^{1+sin{x}}-1 = 1 implies$ $lim_{xrightarrow0} log(2^{1 +sin{x}}-1) = 0$
Therefore:
$lim_{xrightarrow0}frac{1}{2x-1} log(2^{1 +sin{x}}-1) = 0 $ ?
But I do know that it should equal to 2.
How is it so? Where have I done the mistake or how can I achieve the right result?
real-analysis limits logarithms
edited Jan 31 at 12:29
Did
249k23227466
asked Jan 31 at 12:00
cris14cris14
1338
$endgroup$
add a comment |
$begingroup$
Why does $limlimits_{xrightarrow0}frac{1}{2x-1} log(2^{1+sin{x}}-1)= 2 $?
I have:
$lim_{xrightarrow0}frac{1}{2x-1} = -1 $
$lim_{xrightarrow0}2^{1+sin{x}}-1 = 1 implies$ $lim_{xrightarrow0} log(2^{1 +sin{x}}-1) = 0$
Therefore:
$lim_{xrightarrow0}frac{1}{2x-1} log(2^{1 +sin{x}}-1) = 0 $ ?
But I do know that it should equal to 2.
How is it so? Where have I done the mistake or how can I achieve the right result?
real-analysis limits logarithms
edited Jan 31 at 12:29
Did
249k23227466
asked Jan 31 at 12:00
cris14cris14
1338
$endgroup$
$begingroup$
No. Substitute $x = 0$ and you get 0. Heine's definition gives you the immediate answer.
$endgroup$
– PierreCarre
Jan 31 at 12:03
3
$begingroup$
I think it's a mistake in your book.
$endgroup$
– Michael Rozenberg
Jan 31 at 12:05
3
$begingroup$
Did you mistype $2x-1$ for $2^x-1$ by any chance?
$endgroup$
– Did
Jan 31 at 12:30
$begingroup$
Yes, it was typo in the solution.
$endgroup$
– cris14
Jan 31 at 13:08
add a comment |
$begingroup$
Why does $limlimits_{xrightarrow0}frac{1}{2x-1} log(2^{1+sin{x}}-1)= 2 $?
I have:
$lim_{xrightarrow0}frac{1}{2x-1} = -1 $
$lim_{xrightarrow0}2^{1+sin{x}}-1 = 1 implies$ $lim_{xrightarrow0} log(2^{1 +sin{x}}-1) = 0$
Therefore:
$lim_{xrightarrow0}frac{1}{2x-1} log(2^{1 +sin{x}}-1) = 0 $ ?
But I do know that it should equal to 2.
How is it so? Where have I done the mistake or how can I achieve the right result?
real-analysis limits logarithms
edited Jan 31 at 12:29
Did
249k23227466
asked Jan 31 at 12:00
cris14cris14
1338
$endgroup$
Why does $limlimits_{xrightarrow0}frac{1}{2x-1} log(2^{1+sin{x}}-1)= 2 $?
I have:
$lim_{xrightarrow0}frac{1}{2x-1} = -1 $
$lim_{xrightarrow0}2^{1+sin{x}}-1 = 1 implies$ $lim_{xrightarrow0} log(2^{1 +sin{x}}-1) = 0$
Therefore:
$lim_{xrightarrow0}frac{1}{2x-1} log(2^{1 +sin{x}}-1) = 0 $ ?
But I do know that it should equal to 2.
How is it so? Where have I done the mistake or how can I achieve the right result?
real-analysis limits logarithms
real-analysis limits logarithms
edited Jan 31 at 12:29
Did
249k23227466
asked Jan 31 at 12:00
cris14cris14
1338
edited Jan 31 at 12:29
Did
249k23227466
asked Jan 31 at 12:00
cris14cris14
1338
edited Jan 31 at 12:29
Did
249k23227466
edited Jan 31 at 12:29
Did
249k23227466
edited Jan 31 at 12:29
Did
249k23227466
249k23227466
asked Jan 31 at 12:00
cris14cris14
1338
asked Jan 31 at 12:00
cris14cris14
1338
asked Jan 31 at 12:00
cris14cris14
1338
1338
$begingroup$
No. Substitute $x = 0$ and you get 0. Heine's definition gives you the immediate answer.
$endgroup$
– PierreCarre
Jan 31 at 12:03
3
$begingroup$
I think it's a mistake in your book.
$endgroup$
– Michael Rozenberg
Jan 31 at 12:05
3
$begingroup$
Did you mistype $2x-1$ for $2^x-1$ by any chance?
$endgroup$
– Did
Jan 31 at 12:30
$begingroup$
Yes, it was typo in the solution.
$endgroup$
– cris14
Jan 31 at 13:08
add a comment |
$begingroup$
No. Substitute $x = 0$ and you get 0. Heine's definition gives you the immediate answer.
$endgroup$
– PierreCarre
Jan 31 at 12:03
3
$begingroup$
I think it's a mistake in your book.
$endgroup$
– Michael Rozenberg
Jan 31 at 12:05
3
$begingroup$
Did you mistype $2x-1$ for $2^x-1$ by any chance?
$endgroup$
– Did
Jan 31 at 12:30
$begingroup$
Yes, it was typo in the solution.
$endgroup$
– cris14
Jan 31 at 13:08
$begingroup$
No. Substitute $x = 0$ and you get 0. Heine's definition gives you the immediate answer.
$endgroup$
– PierreCarre
Jan 31 at 12:03
$begingroup$
No. Substitute $x = 0$ and you get 0. Heine's definition gives you the immediate answer.
$endgroup$
– PierreCarre
Jan 31 at 12:03
3
3
$begingroup$
I think it's a mistake in your book.
$endgroup$
– Michael Rozenberg
Jan 31 at 12:05
$begingroup$
I think it's a mistake in your book.
$endgroup$
– Michael Rozenberg
Jan 31 at 12:05
3
3
$begingroup$
Did you mistype $2x-1$ for $2^x-1$ by any chance?
$endgroup$
– Did
Jan 31 at 12:30
$begingroup$
Did you mistype $2x-1$ for $2^x-1$ by any chance?
$endgroup$
– Did
Jan 31 at 12:30
$begingroup$
Yes, it was typo in the solution.
$endgroup$
– cris14
Jan 31 at 13:08
$begingroup$
Yes, it was typo in the solution.
$endgroup$
– cris14
Jan 31 at 13:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That's a typo. Most probably it should be
$$lim_{xto 0}frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1} = 2$$
This is indeed true. L'Hospital gives for $xto 0$
begin{eqnarray*} frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1}
& stackrel{L'Hosp.}{sim} & frac{log 2 cdot 2^{1+sin x} cdot cos x}{log 2 cdot 2^x cdot (2^{1+sin x} - 1)}\
& stackrel{xto 0}{longrightarrow} & 2
end{eqnarray*}
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
$endgroup$
$begingroup$
This seems to be the right problem statement (+1)
$endgroup$
– roman
Jan 31 at 12:16
$begingroup$
@roman Should I delete this post then?
$endgroup$
– cris14
Jan 31 at 12:34
1
$begingroup$
@cris14 I think it is good when learners see that their textbooks contain mistakes too :-). So, this sharpens their attention (hopefully).
$endgroup$
– trancelocation
Jan 31 at 12:38
1
$begingroup$
@cris14 Personally I do not see any reason to delete the post.
$endgroup$
– roman
Jan 31 at 12:40
add a comment |
$begingroup$
Is the value of the limit given in the answer section and is equal to $2$? If so I believe it is a mistake. Because:
$$
lim_{xto 0}left(frac{1}{2x-1}log(2^{1+sin x} - 1)right) \
= lim_{xto 0}left(frac{1}{2x-1}right)cdot lim_{xto 0}log(2^{1+sin x} - 1)
$$
Splitting the limit is valid since both limits exist. Now if you substitute $x = 0$ you may obtain:
$$
left({1over 2cdot 0 - 1}right)cdot log(2^{1+0} - 1) = -1cdot log 1 = 0
$$
Therefore your limit is equal to $0$. You may also want to take a look at the graph.
Yet it's not a very formal way to find the limits it may still give you some insights. Clearly the function is crossing the origin at $x=0$ matching the results.
answered Jan 31 at 12:09
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
The function $f(x):=dfrac{1}{2x-1} log(2^{1+sin{x}}-1), quad x in (- pi/2, pi/2)$, is continuous, hence $ lim_{x to 0}f(x)=f(0)=0.$
Your book is wrong ! And you are wright !
answered Jan 31 at 12:13
FredFred
48.6k11849
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
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active
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votes
$begingroup$
That's a typo. Most probably it should be
$$lim_{xto 0}frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1} = 2$$
This is indeed true. L'Hospital gives for $xto 0$
begin{eqnarray*} frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1}
& stackrel{L'Hosp.}{sim} & frac{log 2 cdot 2^{1+sin x} cdot cos x}{log 2 cdot 2^x cdot (2^{1+sin x} - 1)}\
& stackrel{xto 0}{longrightarrow} & 2
end{eqnarray*}
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
$endgroup$
$begingroup$
This seems to be the right problem statement (+1)
$endgroup$
– roman
Jan 31 at 12:16
$begingroup$
@roman Should I delete this post then?
$endgroup$
– cris14
Jan 31 at 12:34
1
$begingroup$
@cris14 I think it is good when learners see that their textbooks contain mistakes too :-). So, this sharpens their attention (hopefully).
$endgroup$
– trancelocation
Jan 31 at 12:38
1
$begingroup$
@cris14 Personally I do not see any reason to delete the post.
$endgroup$
– roman
Jan 31 at 12:40
add a comment |
$begingroup$
That's a typo. Most probably it should be
$$lim_{xto 0}frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1} = 2$$
This is indeed true. L'Hospital gives for $xto 0$
begin{eqnarray*} frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1}
& stackrel{L'Hosp.}{sim} & frac{log 2 cdot 2^{1+sin x} cdot cos x}{log 2 cdot 2^x cdot (2^{1+sin x} - 1)}\
& stackrel{xto 0}{longrightarrow} & 2
end{eqnarray*}
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
$endgroup$
$begingroup$
This seems to be the right problem statement (+1)
$endgroup$
– roman
Jan 31 at 12:16
$begingroup$
@roman Should I delete this post then?
$endgroup$
– cris14
Jan 31 at 12:34
1
$begingroup$
@cris14 I think it is good when learners see that their textbooks contain mistakes too :-). So, this sharpens their attention (hopefully).
$endgroup$
– trancelocation
Jan 31 at 12:38
1
$begingroup$
@cris14 Personally I do not see any reason to delete the post.
$endgroup$
– roman
Jan 31 at 12:40
add a comment |
$begingroup$
That's a typo. Most probably it should be
$$lim_{xto 0}frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1} = 2$$
This is indeed true. L'Hospital gives for $xto 0$
begin{eqnarray*} frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1}
& stackrel{L'Hosp.}{sim} & frac{log 2 cdot 2^{1+sin x} cdot cos x}{log 2 cdot 2^x cdot (2^{1+sin x} - 1)}\
& stackrel{xto 0}{longrightarrow} & 2
end{eqnarray*}
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
$endgroup$
That's a typo. Most probably it should be
$$lim_{xto 0}frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1} = 2$$
This is indeed true. L'Hospital gives for $xto 0$
begin{eqnarray*} frac{log (2^{1+sin x} - 1)}{color{red}{2^x}-1}
& stackrel{L'Hosp.}{sim} & frac{log 2 cdot 2^{1+sin x} cdot cos x}{log 2 cdot 2^x cdot (2^{1+sin x} - 1)}\
& stackrel{xto 0}{longrightarrow} & 2
end{eqnarray*}
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 12:15
trancelocationtrancelocation
13.6k1829
13.6k1829
$begingroup$
This seems to be the right problem statement (+1)
$endgroup$
– roman
Jan 31 at 12:16
$begingroup$
@roman Should I delete this post then?
$endgroup$
– cris14
Jan 31 at 12:34
1
$begingroup$
@cris14 I think it is good when learners see that their textbooks contain mistakes too :-). So, this sharpens their attention (hopefully).
$endgroup$
– trancelocation
Jan 31 at 12:38
1
$begingroup$
@cris14 Personally I do not see any reason to delete the post.
$endgroup$
– roman
Jan 31 at 12:40
add a comment |
$begingroup$
This seems to be the right problem statement (+1)
$endgroup$
– roman
Jan 31 at 12:16
$begingroup$
@roman Should I delete this post then?
$endgroup$
– cris14
Jan 31 at 12:34
1
$begingroup$
@cris14 I think it is good when learners see that their textbooks contain mistakes too :-). So, this sharpens their attention (hopefully).
$endgroup$
– trancelocation
Jan 31 at 12:38
1
$begingroup$
@cris14 Personally I do not see any reason to delete the post.
$endgroup$
– roman
Jan 31 at 12:40
$begingroup$
This seems to be the right problem statement (+1)
$endgroup$
– roman
Jan 31 at 12:16
$begingroup$
This seems to be the right problem statement (+1)
$endgroup$
– roman
Jan 31 at 12:16
$begingroup$
@roman Should I delete this post then?
$endgroup$
– cris14
Jan 31 at 12:34
$begingroup$
@roman Should I delete this post then?
$endgroup$
– cris14
Jan 31 at 12:34
1
1
$begingroup$
@cris14 I think it is good when learners see that their textbooks contain mistakes too :-). So, this sharpens their attention (hopefully).
$endgroup$
– trancelocation
Jan 31 at 12:38
$begingroup$
@cris14 I think it is good when learners see that their textbooks contain mistakes too :-). So, this sharpens their attention (hopefully).
$endgroup$
– trancelocation
Jan 31 at 12:38
1
1
$begingroup$
@cris14 Personally I do not see any reason to delete the post.
$endgroup$
– roman
Jan 31 at 12:40
$begingroup$
@cris14 Personally I do not see any reason to delete the post.
$endgroup$
– roman
Jan 31 at 12:40
add a comment |
$begingroup$
Is the value of the limit given in the answer section and is equal to $2$? If so I believe it is a mistake. Because:
$$
lim_{xto 0}left(frac{1}{2x-1}log(2^{1+sin x} - 1)right) \
= lim_{xto 0}left(frac{1}{2x-1}right)cdot lim_{xto 0}log(2^{1+sin x} - 1)
$$
Splitting the limit is valid since both limits exist. Now if you substitute $x = 0$ you may obtain:
$$
left({1over 2cdot 0 - 1}right)cdot log(2^{1+0} - 1) = -1cdot log 1 = 0
$$
Therefore your limit is equal to $0$. You may also want to take a look at the graph.
Yet it's not a very formal way to find the limits it may still give you some insights. Clearly the function is crossing the origin at $x=0$ matching the results.
answered Jan 31 at 12:09
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
Is the value of the limit given in the answer section and is equal to $2$? If so I believe it is a mistake. Because:
$$
lim_{xto 0}left(frac{1}{2x-1}log(2^{1+sin x} - 1)right) \
= lim_{xto 0}left(frac{1}{2x-1}right)cdot lim_{xto 0}log(2^{1+sin x} - 1)
$$
Splitting the limit is valid since both limits exist. Now if you substitute $x = 0$ you may obtain:
$$
left({1over 2cdot 0 - 1}right)cdot log(2^{1+0} - 1) = -1cdot log 1 = 0
$$
Therefore your limit is equal to $0$. You may also want to take a look at the graph.
Yet it's not a very formal way to find the limits it may still give you some insights. Clearly the function is crossing the origin at $x=0$ matching the results.
answered Jan 31 at 12:09
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
Is the value of the limit given in the answer section and is equal to $2$? If so I believe it is a mistake. Because:
$$
lim_{xto 0}left(frac{1}{2x-1}log(2^{1+sin x} - 1)right) \
= lim_{xto 0}left(frac{1}{2x-1}right)cdot lim_{xto 0}log(2^{1+sin x} - 1)
$$
Splitting the limit is valid since both limits exist. Now if you substitute $x = 0$ you may obtain:
$$
left({1over 2cdot 0 - 1}right)cdot log(2^{1+0} - 1) = -1cdot log 1 = 0
$$
Therefore your limit is equal to $0$. You may also want to take a look at the graph.
Yet it's not a very formal way to find the limits it may still give you some insights. Clearly the function is crossing the origin at $x=0$ matching the results.
answered Jan 31 at 12:09
romanroman
2,43721226
$endgroup$
Is the value of the limit given in the answer section and is equal to $2$? If so I believe it is a mistake. Because:
$$
lim_{xto 0}left(frac{1}{2x-1}log(2^{1+sin x} - 1)right) \
= lim_{xto 0}left(frac{1}{2x-1}right)cdot lim_{xto 0}log(2^{1+sin x} - 1)
$$
Splitting the limit is valid since both limits exist. Now if you substitute $x = 0$ you may obtain:
$$
left({1over 2cdot 0 - 1}right)cdot log(2^{1+0} - 1) = -1cdot log 1 = 0
$$
Therefore your limit is equal to $0$. You may also want to take a look at the graph.
Yet it's not a very formal way to find the limits it may still give you some insights. Clearly the function is crossing the origin at $x=0$ matching the results.
answered Jan 31 at 12:09
romanroman
2,43721226
answered Jan 31 at 12:09
romanroman
2,43721226
answered Jan 31 at 12:09
romanroman
2,43721226
answered Jan 31 at 12:09
romanroman
2,43721226
2,43721226
add a comment |
add a comment |
$begingroup$
The function $f(x):=dfrac{1}{2x-1} log(2^{1+sin{x}}-1), quad x in (- pi/2, pi/2)$, is continuous, hence $ lim_{x to 0}f(x)=f(0)=0.$
Your book is wrong ! And you are wright !
answered Jan 31 at 12:13
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
The function $f(x):=dfrac{1}{2x-1} log(2^{1+sin{x}}-1), quad x in (- pi/2, pi/2)$, is continuous, hence $ lim_{x to 0}f(x)=f(0)=0.$
Your book is wrong ! And you are wright !
answered Jan 31 at 12:13
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
The function $f(x):=dfrac{1}{2x-1} log(2^{1+sin{x}}-1), quad x in (- pi/2, pi/2)$, is continuous, hence $ lim_{x to 0}f(x)=f(0)=0.$
Your book is wrong ! And you are wright !
answered Jan 31 at 12:13
FredFred
48.6k11849
$endgroup$
The function $f(x):=dfrac{1}{2x-1} log(2^{1+sin{x}}-1), quad x in (- pi/2, pi/2)$, is continuous, hence $ lim_{x to 0}f(x)=f(0)=0.$
Your book is wrong ! And you are wright !
answered Jan 31 at 12:13
FredFred
48.6k11849
answered Jan 31 at 12:13
FredFred
48.6k11849
answered Jan 31 at 12:13
FredFred
48.6k11849
answered Jan 31 at 12:13
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
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$begingroup$
No. Substitute $x = 0$ and you get 0. Heine's definition gives you the immediate answer.
$endgroup$
– PierreCarre
Jan 31 at 12:03
3
$begingroup$
I think it's a mistake in your book.
$endgroup$
– Michael Rozenberg
Jan 31 at 12:05
3
$begingroup$
Did you mistype $2x-1$ for $2^x-1$ by any chance?
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– Did
Jan 31 at 12:30
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Yes, it was typo in the solution.
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– cris14
Jan 31 at 13:08