Mixing
Evaluate $lim_{ntoinfty}frac{1}{sin(n)}-frac{1}{n}$ and $lim_{ntoinfty}sum_{k=1}^nfrac{1}{n-kx}$ using...
$begingroup$
I have to calculate the two following limits:
a) $lim_{ntoinfty}frac{1}{sin(n)}-frac{1}{n}$.
b) $lim_{ntoinfty}sum_{k=1}^nfrac{1}{n-kx}$ for $-1<x<1$
Hint : use L'Hospital and Riemann sums.
a) So first I get common denominator $lim_{ntoinfty}frac{n-sin(n)}{nsin(n)}$, then I use L'Hospital $lim_{ntoinfty}frac{1-cos(n)}{sin(n)+ncos(n)}$. Now, if $n$ is not an odd multiple of $frac{pi}{2}$, we get $0$. If it's an odd multiple, we get $pm 1$. Now I'm not sure about my method because WolframAlpha gets another result : https://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+of+1%2Fsin(n)-1%2Fn. Their result is $-infty$ to$-1$, $1$ to $infty$
b) Here I thought about factoring out an $frac{1}{n}$ and making the substitution $x=frac{k}{n}$, so we get
$int_{0}^{1}frac{1}{1-x^2}dx=int_{0}^{1}frac{1}{(1-x)(1+x)}dx$. Now we could make partial fraction decomposition to get $frac{1}{2(x+1)}-frac{1}{2(x-1)}$. And so, if we integrate that, we get $frac{1}{2}log(x+1)-frac{1}{2}log(x-1)$. Now what I find strange is that first we can let $x=frac{k}{n}$ if $x$ is already in the equation. Second, if we evaluate that from $0$ to $1$, we get $frac{1}{2}log(2)+frac{1}{2}log(-1)-frac{1}{2}log(0)$ so I don't know if my approach is correct.
Thanks for your help !
Edit : 
Edit 2: For b), as said, we need to use another variable, so we get $int_0^1frac{1}{1-yx}dy=-frac{log(1-x)}{x}$ which seems valid if, as given $-1<x<1$.
For a), as said, the limit does not exist. They probably meant the limit as n approaches 0. In this case, we can use l'Hospital a second time to get $frac{sin(n)}{cos(n)+cos(n)-nsin(n)}$ which gives $0$ as n approaches zero.
real-analysis calculus limits analysis convergence
asked Jan 17 at 8:55
PoujhPoujh
616516
$endgroup$
add a comment |
$begingroup$
I have to calculate the two following limits:
a) $lim_{ntoinfty}frac{1}{sin(n)}-frac{1}{n}$.
b) $lim_{ntoinfty}sum_{k=1}^nfrac{1}{n-kx}$ for $-1<x<1$
Hint : use L'Hospital and Riemann sums.
a) So first I get common denominator $lim_{ntoinfty}frac{n-sin(n)}{nsin(n)}$, then I use L'Hospital $lim_{ntoinfty}frac{1-cos(n)}{sin(n)+ncos(n)}$. Now, if $n$ is not an odd multiple of $frac{pi}{2}$, we get $0$. If it's an odd multiple, we get $pm 1$. Now I'm not sure about my method because WolframAlpha gets another result : https://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+of+1%2Fsin(n)-1%2Fn. Their result is $-infty$ to$-1$, $1$ to $infty$
b) Here I thought about factoring out an $frac{1}{n}$ and making the substitution $x=frac{k}{n}$, so we get
$int_{0}^{1}frac{1}{1-x^2}dx=int_{0}^{1}frac{1}{(1-x)(1+x)}dx$. Now we could make partial fraction decomposition to get $frac{1}{2(x+1)}-frac{1}{2(x-1)}$. And so, if we integrate that, we get $frac{1}{2}log(x+1)-frac{1}{2}log(x-1)$. Now what I find strange is that first we can let $x=frac{k}{n}$ if $x$ is already in the equation. Second, if we evaluate that from $0$ to $1$, we get $frac{1}{2}log(2)+frac{1}{2}log(-1)-frac{1}{2}log(0)$ so I don't know if my approach is correct.
Thanks for your help !
Edit :
Edit 2: For b), as said, we need to use another variable, so we get $int_0^1frac{1}{1-yx}dy=-frac{log(1-x)}{x}$ which seems valid if, as given $-1<x<1$.
For a), as said, the limit does not exist. They probably meant the limit as n approaches 0. In this case, we can use l'Hospital a second time to get $frac{sin(n)}{cos(n)+cos(n)-nsin(n)}$ which gives $0$ as n approaches zero.
real-analysis calculus limits analysis convergence
asked Jan 17 at 8:55
PoujhPoujh
616516
$endgroup$
$begingroup$
For part b): call $k/n=y$, you will see your mistake.
$endgroup$
– lcv
Jan 17 at 9:08
add a comment |
$begingroup$
I have to calculate the two following limits:
a) $lim_{ntoinfty}frac{1}{sin(n)}-frac{1}{n}$.
b) $lim_{ntoinfty}sum_{k=1}^nfrac{1}{n-kx}$ for $-1<x<1$
Hint : use L'Hospital and Riemann sums.
a) So first I get common denominator $lim_{ntoinfty}frac{n-sin(n)}{nsin(n)}$, then I use L'Hospital $lim_{ntoinfty}frac{1-cos(n)}{sin(n)+ncos(n)}$. Now, if $n$ is not an odd multiple of $frac{pi}{2}$, we get $0$. If it's an odd multiple, we get $pm 1$. Now I'm not sure about my method because WolframAlpha gets another result : https://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+of+1%2Fsin(n)-1%2Fn. Their result is $-infty$ to$-1$, $1$ to $infty$
b) Here I thought about factoring out an $frac{1}{n}$ and making the substitution $x=frac{k}{n}$, so we get
$int_{0}^{1}frac{1}{1-x^2}dx=int_{0}^{1}frac{1}{(1-x)(1+x)}dx$. Now we could make partial fraction decomposition to get $frac{1}{2(x+1)}-frac{1}{2(x-1)}$. And so, if we integrate that, we get $frac{1}{2}log(x+1)-frac{1}{2}log(x-1)$. Now what I find strange is that first we can let $x=frac{k}{n}$ if $x$ is already in the equation. Second, if we evaluate that from $0$ to $1$, we get $frac{1}{2}log(2)+frac{1}{2}log(-1)-frac{1}{2}log(0)$ so I don't know if my approach is correct.
Thanks for your help !
Edit :
Edit 2: For b), as said, we need to use another variable, so we get $int_0^1frac{1}{1-yx}dy=-frac{log(1-x)}{x}$ which seems valid if, as given $-1<x<1$.
For a), as said, the limit does not exist. They probably meant the limit as n approaches 0. In this case, we can use l'Hospital a second time to get $frac{sin(n)}{cos(n)+cos(n)-nsin(n)}$ which gives $0$ as n approaches zero.
real-analysis calculus limits analysis convergence
asked Jan 17 at 8:55
PoujhPoujh
616516
$endgroup$
I have to calculate the two following limits:
a) $lim_{ntoinfty}frac{1}{sin(n)}-frac{1}{n}$.
b) $lim_{ntoinfty}sum_{k=1}^nfrac{1}{n-kx}$ for $-1<x<1$
Hint : use L'Hospital and Riemann sums.
a) So first I get common denominator $lim_{ntoinfty}frac{n-sin(n)}{nsin(n)}$, then I use L'Hospital $lim_{ntoinfty}frac{1-cos(n)}{sin(n)+ncos(n)}$. Now, if $n$ is not an odd multiple of $frac{pi}{2}$, we get $0$. If it's an odd multiple, we get $pm 1$. Now I'm not sure about my method because WolframAlpha gets another result : https://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+of+1%2Fsin(n)-1%2Fn. Their result is $-infty$ to$-1$, $1$ to $infty$
b) Here I thought about factoring out an $frac{1}{n}$ and making the substitution $x=frac{k}{n}$, so we get
$int_{0}^{1}frac{1}{1-x^2}dx=int_{0}^{1}frac{1}{(1-x)(1+x)}dx$. Now we could make partial fraction decomposition to get $frac{1}{2(x+1)}-frac{1}{2(x-1)}$. And so, if we integrate that, we get $frac{1}{2}log(x+1)-frac{1}{2}log(x-1)$. Now what I find strange is that first we can let $x=frac{k}{n}$ if $x$ is already in the equation. Second, if we evaluate that from $0$ to $1$, we get $frac{1}{2}log(2)+frac{1}{2}log(-1)-frac{1}{2}log(0)$ so I don't know if my approach is correct.
Thanks for your help !
Edit :
Edit 2: For b), as said, we need to use another variable, so we get $int_0^1frac{1}{1-yx}dy=-frac{log(1-x)}{x}$ which seems valid if, as given $-1<x<1$.
For a), as said, the limit does not exist. They probably meant the limit as n approaches 0. In this case, we can use l'Hospital a second time to get $frac{sin(n)}{cos(n)+cos(n)-nsin(n)}$ which gives $0$ as n approaches zero.
real-analysis calculus limits analysis convergence
real-analysis calculus limits analysis convergence
asked Jan 17 at 8:55
PoujhPoujh
616516
asked Jan 17 at 8:55
PoujhPoujh
616516
edited Jan 17 at 10:02
Poujh
asked Jan 17 at 8:55
PoujhPoujh
616516
asked Jan 17 at 8:55
PoujhPoujh
616516
asked Jan 17 at 8:55
PoujhPoujh
616516
616516
$begingroup$
For part b): call $k/n=y$, you will see your mistake.
$endgroup$
– lcv
Jan 17 at 9:08
add a comment |
$begingroup$
For part b): call $k/n=y$, you will see your mistake.
$endgroup$
– lcv
Jan 17 at 9:08
$begingroup$
For part b): call $k/n=y$, you will see your mistake.
$endgroup$
– lcv
Jan 17 at 9:08
$begingroup$
For part b): call $k/n=y$, you will see your mistake.
$endgroup$
– lcv
Jan 17 at 9:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
- Technically speaking, it is illegal to use L'Hopital rule to sequential limits. And I don't think such limit exists. Since the hint is the L'Hopital rule, I think it is more likely to be
$$
lim_{x to 0} frac 1{sin x} - frac 1x.
$$
To let the limit be nonzero, maybe it also could be
$$
lim_{x to 0} frac 1{sin^2 x} - frac 1{x^2}.
$$
- You got the letters wrong. $x$ is a given constant. To write Riemann sum you should consider the function
$$
f(t) = frac 1{1 - x t}, t in [0,1].
$$
UPDATE
If you insist, then such limit does not exist.
Proof. Assume such limit exists, let it be $A$, then using the arithmetic operation of limits,
$$
lim_{n to infty} frac 1{sin n} = lim_{n to infty} frac 1{sin n} - frac 1n + lim_{n to infty} frac 1n = A + 0 = A.
$$
Easy to see that $A neq 0$ because
$$
leftvert frac 1 {sin n} right vert geqslant 1.
$$
Then using the arithmetic operation again,
$$
lim_{nto infty} sin n = frac 1A
$$
exists. But in fact $sin n$ has no limits [proof omitted, if you need then I will add], contradiction. Hence the limit does not exist. $square$
answered Jan 17 at 9:12
xbhxbh
6,2151522
$endgroup$
$begingroup$
Well, I have this old exam in front of me and they clearly state $lim_{nto infty}$ for a) I think I will add a picture in my question.
$endgroup$
– Poujh
Jan 17 at 9:16
$begingroup$
See edit with picture
$endgroup$
– Poujh
Jan 17 at 9:18
$begingroup$
Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012
$endgroup$
– Poujh
Jan 17 at 9:20
$begingroup$
@Poujh I don't know…… And I cannot possibly know.
$endgroup$
– xbh
Jan 17 at 9:32
$begingroup$
Yeah, I know, but I don't know too haha. I'm just making suppositions.
$endgroup$
– Poujh
Jan 17 at 9:33
add a comment |
$begingroup$
a) This limit (obviously) does not exist. Are you sure it is correctly stated?
b) Your approach seems right, but if x is already in there, you need to call your dummy variable different, of course. As a consequence you will get an answer depending on x.
answered Jan 17 at 9:06
KlausKlaus
1,8179
$endgroup$
$begingroup$
a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get
$endgroup$
– Poujh
Jan 17 at 9:10
$begingroup$
For b), I get $frac{-log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic
$endgroup$
– Poujh
Jan 17 at 9:15
$begingroup$
See my picture in the edit
$endgroup$
– Poujh
Jan 17 at 9:19
1
$begingroup$
Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $frac{-log(1-x)}{x}$. For (a) they probably made a mistake, $n to 0$ would make more sense.
$endgroup$
– Klaus
Jan 17 at 9:26
$begingroup$
Well that's why I said you lost a minus sign sowhere. ;-)
$endgroup$
– Klaus
Jan 17 at 9:28
|
show 2 more comments
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- Technically speaking, it is illegal to use L'Hopital rule to sequential limits. And I don't think such limit exists. Since the hint is the L'Hopital rule, I think it is more likely to be
$$
lim_{x to 0} frac 1{sin x} - frac 1x.
$$
To let the limit be nonzero, maybe it also could be
$$
lim_{x to 0} frac 1{sin^2 x} - frac 1{x^2}.
$$
- You got the letters wrong. $x$ is a given constant. To write Riemann sum you should consider the function
$$
f(t) = frac 1{1 - x t}, t in [0,1].
$$
UPDATE
If you insist, then such limit does not exist.
Proof. Assume such limit exists, let it be $A$, then using the arithmetic operation of limits,
$$
lim_{n to infty} frac 1{sin n} = lim_{n to infty} frac 1{sin n} - frac 1n + lim_{n to infty} frac 1n = A + 0 = A.
$$
Easy to see that $A neq 0$ because
$$
leftvert frac 1 {sin n} right vert geqslant 1.
$$
Then using the arithmetic operation again,
$$
lim_{nto infty} sin n = frac 1A
$$
exists. But in fact $sin n$ has no limits [proof omitted, if you need then I will add], contradiction. Hence the limit does not exist. $square$
answered Jan 17 at 9:12
xbhxbh
6,2151522
$endgroup$
$begingroup$
Well, I have this old exam in front of me and they clearly state $lim_{nto infty}$ for a) I think I will add a picture in my question.
$endgroup$
– Poujh
Jan 17 at 9:16
$begingroup$
See edit with picture
$endgroup$
– Poujh
Jan 17 at 9:18
$begingroup$
Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012
$endgroup$
– Poujh
Jan 17 at 9:20
$begingroup$
@Poujh I don't know…… And I cannot possibly know.
$endgroup$
– xbh
Jan 17 at 9:32
$begingroup$
Yeah, I know, but I don't know too haha. I'm just making suppositions.
$endgroup$
– Poujh
Jan 17 at 9:33
add a comment |
$begingroup$
- Technically speaking, it is illegal to use L'Hopital rule to sequential limits. And I don't think such limit exists. Since the hint is the L'Hopital rule, I think it is more likely to be
$$
lim_{x to 0} frac 1{sin x} - frac 1x.
$$
To let the limit be nonzero, maybe it also could be
$$
lim_{x to 0} frac 1{sin^2 x} - frac 1{x^2}.
$$
- You got the letters wrong. $x$ is a given constant. To write Riemann sum you should consider the function
$$
f(t) = frac 1{1 - x t}, t in [0,1].
$$
UPDATE
If you insist, then such limit does not exist.
Proof. Assume such limit exists, let it be $A$, then using the arithmetic operation of limits,
$$
lim_{n to infty} frac 1{sin n} = lim_{n to infty} frac 1{sin n} - frac 1n + lim_{n to infty} frac 1n = A + 0 = A.
$$
Easy to see that $A neq 0$ because
$$
leftvert frac 1 {sin n} right vert geqslant 1.
$$
Then using the arithmetic operation again,
$$
lim_{nto infty} sin n = frac 1A
$$
exists. But in fact $sin n$ has no limits [proof omitted, if you need then I will add], contradiction. Hence the limit does not exist. $square$
answered Jan 17 at 9:12
xbhxbh
6,2151522
$endgroup$
$begingroup$
Well, I have this old exam in front of me and they clearly state $lim_{nto infty}$ for a) I think I will add a picture in my question.
$endgroup$
– Poujh
Jan 17 at 9:16
$begingroup$
See edit with picture
$endgroup$
– Poujh
Jan 17 at 9:18
$begingroup$
Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012
$endgroup$
– Poujh
Jan 17 at 9:20
$begingroup$
@Poujh I don't know…… And I cannot possibly know.
$endgroup$
– xbh
Jan 17 at 9:32
$begingroup$
Yeah, I know, but I don't know too haha. I'm just making suppositions.
$endgroup$
– Poujh
Jan 17 at 9:33
add a comment |
$begingroup$
- Technically speaking, it is illegal to use L'Hopital rule to sequential limits. And I don't think such limit exists. Since the hint is the L'Hopital rule, I think it is more likely to be
$$
lim_{x to 0} frac 1{sin x} - frac 1x.
$$
To let the limit be nonzero, maybe it also could be
$$
lim_{x to 0} frac 1{sin^2 x} - frac 1{x^2}.
$$
- You got the letters wrong. $x$ is a given constant. To write Riemann sum you should consider the function
$$
f(t) = frac 1{1 - x t}, t in [0,1].
$$
UPDATE
If you insist, then such limit does not exist.
Proof. Assume such limit exists, let it be $A$, then using the arithmetic operation of limits,
$$
lim_{n to infty} frac 1{sin n} = lim_{n to infty} frac 1{sin n} - frac 1n + lim_{n to infty} frac 1n = A + 0 = A.
$$
Easy to see that $A neq 0$ because
$$
leftvert frac 1 {sin n} right vert geqslant 1.
$$
Then using the arithmetic operation again,
$$
lim_{nto infty} sin n = frac 1A
$$
exists. But in fact $sin n$ has no limits [proof omitted, if you need then I will add], contradiction. Hence the limit does not exist. $square$
answered Jan 17 at 9:12
xbhxbh
6,2151522
$endgroup$
- Technically speaking, it is illegal to use L'Hopital rule to sequential limits. And I don't think such limit exists. Since the hint is the L'Hopital rule, I think it is more likely to be
$$
lim_{x to 0} frac 1{sin x} - frac 1x.
$$
To let the limit be nonzero, maybe it also could be
$$
lim_{x to 0} frac 1{sin^2 x} - frac 1{x^2}.
$$
- You got the letters wrong. $x$ is a given constant. To write Riemann sum you should consider the function
$$
f(t) = frac 1{1 - x t}, t in [0,1].
$$
UPDATE
If you insist, then such limit does not exist.
Proof. Assume such limit exists, let it be $A$, then using the arithmetic operation of limits,
$$
lim_{n to infty} frac 1{sin n} = lim_{n to infty} frac 1{sin n} - frac 1n + lim_{n to infty} frac 1n = A + 0 = A.
$$
Easy to see that $A neq 0$ because
$$
leftvert frac 1 {sin n} right vert geqslant 1.
$$
Then using the arithmetic operation again,
$$
lim_{nto infty} sin n = frac 1A
$$
exists. But in fact $sin n$ has no limits [proof omitted, if you need then I will add], contradiction. Hence the limit does not exist. $square$
answered Jan 17 at 9:12
xbhxbh
6,2151522
edited Jan 17 at 9:35
answered Jan 17 at 9:12
xbhxbh
6,2151522
answered Jan 17 at 9:12
xbhxbh
6,2151522
answered Jan 17 at 9:12
xbhxbh
6,2151522
6,2151522
$begingroup$
Well, I have this old exam in front of me and they clearly state $lim_{nto infty}$ for a) I think I will add a picture in my question.
$endgroup$
– Poujh
Jan 17 at 9:16
$begingroup$
See edit with picture
$endgroup$
– Poujh
Jan 17 at 9:18
$begingroup$
Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012
$endgroup$
– Poujh
Jan 17 at 9:20
$begingroup$
@Poujh I don't know…… And I cannot possibly know.
$endgroup$
– xbh
Jan 17 at 9:32
$begingroup$
Yeah, I know, but I don't know too haha. I'm just making suppositions.
$endgroup$
– Poujh
Jan 17 at 9:33
add a comment |
$begingroup$
Well, I have this old exam in front of me and they clearly state $lim_{nto infty}$ for a) I think I will add a picture in my question.
$endgroup$
– Poujh
Jan 17 at 9:16
$begingroup$
See edit with picture
$endgroup$
– Poujh
Jan 17 at 9:18
$begingroup$
Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012
$endgroup$
– Poujh
Jan 17 at 9:20
$begingroup$
@Poujh I don't know…… And I cannot possibly know.
$endgroup$
– xbh
Jan 17 at 9:32
$begingroup$
Yeah, I know, but I don't know too haha. I'm just making suppositions.
$endgroup$
– Poujh
Jan 17 at 9:33
$begingroup$
Well, I have this old exam in front of me and they clearly state $lim_{nto infty}$ for a) I think I will add a picture in my question.
$endgroup$
– Poujh
Jan 17 at 9:16
$begingroup$
Well, I have this old exam in front of me and they clearly state $lim_{nto infty}$ for a) I think I will add a picture in my question.
$endgroup$
– Poujh
Jan 17 at 9:16
$begingroup$
See edit with picture
$endgroup$
– Poujh
Jan 17 at 9:18
$begingroup$
See edit with picture
$endgroup$
– Poujh
Jan 17 at 9:18
$begingroup$
Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012
$endgroup$
– Poujh
Jan 17 at 9:20
$begingroup$
Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012
$endgroup$
– Poujh
Jan 17 at 9:20
$begingroup$
@Poujh I don't know…… And I cannot possibly know.
$endgroup$
– xbh
Jan 17 at 9:32
$begingroup$
@Poujh I don't know…… And I cannot possibly know.
$endgroup$
– xbh
Jan 17 at 9:32
$begingroup$
Yeah, I know, but I don't know too haha. I'm just making suppositions.
$endgroup$
– Poujh
Jan 17 at 9:33
$begingroup$
Yeah, I know, but I don't know too haha. I'm just making suppositions.
$endgroup$
– Poujh
Jan 17 at 9:33
add a comment |
$begingroup$
a) This limit (obviously) does not exist. Are you sure it is correctly stated?
b) Your approach seems right, but if x is already in there, you need to call your dummy variable different, of course. As a consequence you will get an answer depending on x.
answered Jan 17 at 9:06
KlausKlaus
1,8179
$endgroup$
$begingroup$
a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get
$endgroup$
– Poujh
Jan 17 at 9:10
$begingroup$
For b), I get $frac{-log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic
$endgroup$
– Poujh
Jan 17 at 9:15
$begingroup$
See my picture in the edit
$endgroup$
– Poujh
Jan 17 at 9:19
1
$begingroup$
Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $frac{-log(1-x)}{x}$. For (a) they probably made a mistake, $n to 0$ would make more sense.
$endgroup$
– Klaus
Jan 17 at 9:26
$begingroup$
Well that's why I said you lost a minus sign sowhere. ;-)
$endgroup$
– Klaus
Jan 17 at 9:28
|
show 2 more comments
$begingroup$
a) This limit (obviously) does not exist. Are you sure it is correctly stated?
b) Your approach seems right, but if x is already in there, you need to call your dummy variable different, of course. As a consequence you will get an answer depending on x.
answered Jan 17 at 9:06
KlausKlaus
1,8179
$endgroup$
$begingroup$
a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get
$endgroup$
– Poujh
Jan 17 at 9:10
$begingroup$
For b), I get $frac{-log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic
$endgroup$
– Poujh
Jan 17 at 9:15
$begingroup$
See my picture in the edit
$endgroup$
– Poujh
Jan 17 at 9:19
1
$begingroup$
Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $frac{-log(1-x)}{x}$. For (a) they probably made a mistake, $n to 0$ would make more sense.
$endgroup$
– Klaus
Jan 17 at 9:26
$begingroup$
Well that's why I said you lost a minus sign sowhere. ;-)
$endgroup$
– Klaus
Jan 17 at 9:28
|
show 2 more comments
$begingroup$
a) This limit (obviously) does not exist. Are you sure it is correctly stated?
b) Your approach seems right, but if x is already in there, you need to call your dummy variable different, of course. As a consequence you will get an answer depending on x.
answered Jan 17 at 9:06
KlausKlaus
1,8179
$endgroup$
a) This limit (obviously) does not exist. Are you sure it is correctly stated?
b) Your approach seems right, but if x is already in there, you need to call your dummy variable different, of course. As a consequence you will get an answer depending on x.
answered Jan 17 at 9:06
KlausKlaus
1,8179
answered Jan 17 at 9:06
KlausKlaus
1,8179
answered Jan 17 at 9:06
KlausKlaus
1,8179
answered Jan 17 at 9:06
KlausKlaus
1,8179
1,8179
$begingroup$
a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get
$endgroup$
– Poujh
Jan 17 at 9:10
$begingroup$
For b), I get $frac{-log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic
$endgroup$
– Poujh
Jan 17 at 9:15
$begingroup$
See my picture in the edit
$endgroup$
– Poujh
Jan 17 at 9:19
1
$begingroup$
Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $frac{-log(1-x)}{x}$. For (a) they probably made a mistake, $n to 0$ would make more sense.
$endgroup$
– Klaus
Jan 17 at 9:26
$begingroup$
Well that's why I said you lost a minus sign sowhere. ;-)
$endgroup$
– Klaus
Jan 17 at 9:28
|
show 2 more comments
$begingroup$
a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get
$endgroup$
– Poujh
Jan 17 at 9:10
$begingroup$
For b), I get $frac{-log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic
$endgroup$
– Poujh
Jan 17 at 9:15
$begingroup$
See my picture in the edit
$endgroup$
– Poujh
Jan 17 at 9:19
1
$begingroup$
Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $frac{-log(1-x)}{x}$. For (a) they probably made a mistake, $n to 0$ would make more sense.
$endgroup$
– Klaus
Jan 17 at 9:26
$begingroup$
Well that's why I said you lost a minus sign sowhere. ;-)
$endgroup$
– Klaus
Jan 17 at 9:28
$begingroup$
a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get
$endgroup$
– Poujh
Jan 17 at 9:10
$begingroup$
a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get
$endgroup$
– Poujh
Jan 17 at 9:10
$begingroup$
For b), I get $frac{-log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic
$endgroup$
– Poujh
Jan 17 at 9:15
$begingroup$
For b), I get $frac{-log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic
$endgroup$
– Poujh
Jan 17 at 9:15
$begingroup$
See my picture in the edit
$endgroup$
– Poujh
Jan 17 at 9:19
$begingroup$
See my picture in the edit
$endgroup$
– Poujh
Jan 17 at 9:19
1
1
$begingroup$
Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $frac{-log(1-x)}{x}$. For (a) they probably made a mistake, $n to 0$ would make more sense.
$endgroup$
– Klaus
Jan 17 at 9:26
$begingroup$
Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $frac{-log(1-x)}{x}$. For (a) they probably made a mistake, $n to 0$ would make more sense.
$endgroup$
– Klaus
Jan 17 at 9:26
$begingroup$
Well that's why I said you lost a minus sign sowhere. ;-)
$endgroup$
– Klaus
Jan 17 at 9:28
$begingroup$
Well that's why I said you lost a minus sign sowhere. ;-)
$endgroup$
– Klaus
Jan 17 at 9:28
|
show 2 more comments
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$begingroup$
For part b): call $k/n=y$, you will see your mistake.
$endgroup$
– lcv
Jan 17 at 9:08