Mixing
Proof of L'Hopital's rule ($x_0 notin D$)
0
$begingroup$
Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)
We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.
Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.
$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)
If we let $n to infty$ we get
$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$
$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$
calculus limits derivatives
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
$endgroup$
add a comment |
0
$begingroup$
Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)
We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.
Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.
$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)
If we let $n to infty$ we get
$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$
$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$
calculus limits derivatives
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
$endgroup$
$begingroup$
You probably mean L'Hopital's rule
$endgroup$
– roman
Jan 3 at 8:39
$begingroup$
@roman: L'Hôpital's rule, to be precise :)
$endgroup$
– Martin R
Jan 3 at 8:43
$begingroup$
How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
$endgroup$
– 5xum
Jan 3 at 8:47
$begingroup$
@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:50
add a comment |
0
0
0
$begingroup$
Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)
We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.
Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.
$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)
If we let $n to infty$ we get
$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$
$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$
calculus limits derivatives
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
$endgroup$
Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)
We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.
Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.
$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)
If we let $n to infty$ we get
$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$
$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$
calculus limits derivatives
calculus limits derivatives
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
edited Jan 3 at 8:42
Francesco Andreuzzi
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
asked Jan 3 at 8:36
Francesco AndreuzziFrancesco Andreuzzi
63
63
$begingroup$
You probably mean L'Hopital's rule
$endgroup$
– roman
Jan 3 at 8:39
$begingroup$
@roman: L'Hôpital's rule, to be precise :)
$endgroup$
– Martin R
Jan 3 at 8:43
$begingroup$
How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
$endgroup$
– 5xum
Jan 3 at 8:47
$begingroup$
@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:50
add a comment |
$begingroup$
You probably mean L'Hopital's rule
$endgroup$
– roman
Jan 3 at 8:39
$begingroup$
@roman: L'Hôpital's rule, to be precise :)
$endgroup$
– Martin R
Jan 3 at 8:43
$begingroup$
How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
$endgroup$
– 5xum
Jan 3 at 8:47
$begingroup$
@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:50
$begingroup$
You probably mean L'Hopital's rule
$endgroup$
– roman
Jan 3 at 8:39
$begingroup$
You probably mean L'Hopital's rule
$endgroup$
– roman
Jan 3 at 8:39
$begingroup$
@roman: L'Hôpital's rule, to be precise :)
$endgroup$
– Martin R
Jan 3 at 8:43
$begingroup$
@roman: L'Hôpital's rule, to be precise :)
$endgroup$
– Martin R
Jan 3 at 8:43
$begingroup$
How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
$endgroup$
– 5xum
Jan 3 at 8:47
$begingroup$
How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
$endgroup$
– 5xum
Jan 3 at 8:47
$begingroup$
@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:50
$begingroup$
@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:50
add a comment |
1 Answer
1
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oldest
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The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:
- You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.
- You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?
answered Jan 3 at 8:51
5xum5xum
90.1k393161
$endgroup$
$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53
$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54
$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55
$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56
1
$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03
|
show 1 more comment
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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1
$begingroup$
The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:
- You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.
- You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?
answered Jan 3 at 8:51
5xum5xum
90.1k393161
$endgroup$
$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53
$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54
$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55
$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56
1
$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03
|
show 1 more comment
1
$begingroup$
The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:
- You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.
- You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?
answered Jan 3 at 8:51
5xum5xum
90.1k393161
$endgroup$
$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53
$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54
$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55
$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56
1
$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03
|
show 1 more comment
1
1
1
$begingroup$
The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:
- You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.
- You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?
answered Jan 3 at 8:51
5xum5xum
90.1k393161
$endgroup$
The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:
- You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.
- You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?
answered Jan 3 at 8:51
5xum5xum
90.1k393161
answered Jan 3 at 8:51
5xum5xum
90.1k393161
answered Jan 3 at 8:51
5xum5xum
90.1k393161
answered Jan 3 at 8:51
5xum5xum
90.1k393161
90.1k393161
$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53
$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54
$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55
$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56
1
$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03
|
show 1 more comment
$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53
$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54
$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55
$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56
1
$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03
$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53
$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53
$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54
$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54
$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55
$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55
$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56
$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56
1
1
$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03
$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03
|
show 1 more comment
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You probably mean L'Hopital's rule
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– roman
Jan 3 at 8:39
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@roman: L'Hôpital's rule, to be precise :)
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– Martin R
Jan 3 at 8:43
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How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
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– 5xum
Jan 3 at 8:47
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@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
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– Francesco Andreuzzi
Jan 3 at 8:50