Theorem 5.13 in Baby Rudin: Is this statement of the L'Hospital's rule the most optimal one?
Here is Theorem 5.13 (L'Hospital's Rule) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $f$ and $g$ are real and differentiable in $(a, b)$, and $g^prime(x) neq 0$ for all $x in (a, b)$, where $-infty leq a < b leq +infty$. Suppose
$$ frac{f^prime(x)}{g^prime(x)} to A mbox{ as } x to a. tag{13} $$
If
$$ f(x) to 0 mbox{ and } g(x) to 0 mbox{ as } x to a, tag{14} $$
or if
$$ g(x) to +infty mbox{ as } x to a, tag{15} $$
then
$$ frac{f(x)}{g(x)} to A mbox{ as } x to a. tag{16}$$
The analogous statement is of course also true if $x to b$, or if $g(x) to -infty$ in (15). Let us note that we now use the limit concept in the extended sense of Definition 4.33.
Here is Definition 4.33:
Let $f$ be a real function defined on $E subset mathbb{R}$. We say that
$$ f(t) to A mbox{ as } t to x, $$
where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V cap E$ is not empty, and such that $f(t) in U$ for all $t in V cap E$, $t neq x$.
And, here is Rudin's proof:
We first consider the case in which $-infty leq A < +infty$. Choose a real number $q$ such that $A < q$, and then choose $r$ such that $A < r < q$. By (13) there is a point $c in (a, b)$ such that $a < x < c$ implies
$$ frac{ f^prime(x) }{ g^prime(x) } < r. tag{17} $$
If $a < x < y < c$, then Theorem 5.9 shows that there is a point $t in (x, y)$ such that
$$ frac{ f(x)-f(y) }{ g(x)-g(y) } = frac{f^prime(t)}{g^prime(t)} < r. tag{18} $$
Suppose (14) holds. Letting $x to a$ in (18), we see that
$$ frac{f(y)}{g(y)} leq r < q qquad qquad qquad (a < y < c) tag{19} $$
Next, suppose (15) holds. Keeping $y$ fixed in (18), we can choos a point $c_1 in (a, y)$ such that $g(x) > g(y)$ and $g(x) > 0$ if $a < x < c_1$. Multiplying (18) by $left[ g(x)- g(y) right]/g(x)$, we obtain
$$ frac{ f(x) }{ g(x) } < r - r frac{ g(y) }{g(x)} + frac{f(y)}{g(x)} qquad qquad qquad (a < x < c_1). tag{20}$$
If we let $x to a$ in (20), (15) shows that there is a point $c_2 in left( a, c_1 right)$ such that
$$ frac{ f(x) }{ g(x) } < q qquad qquad qquad (a < x < c_2 ). tag{21} $$
Summing up, (19) and (21) show that for any $q$, subject only to the condition $A < q$, there is a point $c_2$ such that $f(x)/g(x) < q$ if $a < x < c_2$.
In the same manner, if $-infty < A leq +infty$, and $p$ is chosen so that $p < A$, we can find a point $c_3$ such that
$$ p < frac{ f(x) }{ g(x) } qquad qquad qquad ( a< x < c_3), tag{22} $$
and (16) follows from these two statements.
Now I have the following queries:
In (20), why has the term $f(x)/g(x)$ not been affected as we let $x to a$, although two of the three terms on the right side have gone to $0$? Don't we need any assumption about $f$ in (15)?
calculus real-analysis analysis derivatives
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
add a comment |
Here is Theorem 5.13 (L'Hospital's Rule) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $f$ and $g$ are real and differentiable in $(a, b)$, and $g^prime(x) neq 0$ for all $x in (a, b)$, where $-infty leq a < b leq +infty$. Suppose
$$ frac{f^prime(x)}{g^prime(x)} to A mbox{ as } x to a. tag{13} $$
If
$$ f(x) to 0 mbox{ and } g(x) to 0 mbox{ as } x to a, tag{14} $$
or if
$$ g(x) to +infty mbox{ as } x to a, tag{15} $$
then
$$ frac{f(x)}{g(x)} to A mbox{ as } x to a. tag{16}$$
The analogous statement is of course also true if $x to b$, or if $g(x) to -infty$ in (15). Let us note that we now use the limit concept in the extended sense of Definition 4.33.
Here is Definition 4.33:
Let $f$ be a real function defined on $E subset mathbb{R}$. We say that
$$ f(t) to A mbox{ as } t to x, $$
where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V cap E$ is not empty, and such that $f(t) in U$ for all $t in V cap E$, $t neq x$.
And, here is Rudin's proof:
We first consider the case in which $-infty leq A < +infty$. Choose a real number $q$ such that $A < q$, and then choose $r$ such that $A < r < q$. By (13) there is a point $c in (a, b)$ such that $a < x < c$ implies
$$ frac{ f^prime(x) }{ g^prime(x) } < r. tag{17} $$
If $a < x < y < c$, then Theorem 5.9 shows that there is a point $t in (x, y)$ such that
$$ frac{ f(x)-f(y) }{ g(x)-g(y) } = frac{f^prime(t)}{g^prime(t)} < r. tag{18} $$
Suppose (14) holds. Letting $x to a$ in (18), we see that
$$ frac{f(y)}{g(y)} leq r < q qquad qquad qquad (a < y < c) tag{19} $$
Next, suppose (15) holds. Keeping $y$ fixed in (18), we can choos a point $c_1 in (a, y)$ such that $g(x) > g(y)$ and $g(x) > 0$ if $a < x < c_1$. Multiplying (18) by $left[ g(x)- g(y) right]/g(x)$, we obtain
$$ frac{ f(x) }{ g(x) } < r - r frac{ g(y) }{g(x)} + frac{f(y)}{g(x)} qquad qquad qquad (a < x < c_1). tag{20}$$
If we let $x to a$ in (20), (15) shows that there is a point $c_2 in left( a, c_1 right)$ such that
$$ frac{ f(x) }{ g(x) } < q qquad qquad qquad (a < x < c_2 ). tag{21} $$
Summing up, (19) and (21) show that for any $q$, subject only to the condition $A < q$, there is a point $c_2$ such that $f(x)/g(x) < q$ if $a < x < c_2$.
In the same manner, if $-infty < A leq +infty$, and $p$ is chosen so that $p < A$, we can find a point $c_3$ such that
$$ p < frac{ f(x) }{ g(x) } qquad qquad qquad ( a< x < c_3), tag{22} $$
and (16) follows from these two statements.
Now I have the following queries:
In (20), why has the term $f(x)/g(x)$ not been affected as we let $x to a$, although two of the three terms on the right side have gone to $0$? Don't we need any assumption about $f$ in (15)?
calculus real-analysis analysis derivatives
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
It is rigorous but terse. We have $f(y) -rg(y)$ fixed and $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} to 0$ as $x to a$, since $g(x) to +infty$. For $x$ sufficiently close to $a$ we have $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} < q -r$ and $f(x)/g(x) < q$.
– RRL
Apr 9 '17 at 0:09
Also, nothing about the limit of $f(x)$ or even its existence is needed to show this.
– RRL
Apr 9 '17 at 0:10
Why is Math SE voting for the closure of this question, I wonder? Is there anything about that is against the etiquette of this wonderful forum?
– Saaqib Mahmood
Apr 9 '17 at 8:33
add a comment |
Here is Theorem 5.13 (L'Hospital's Rule) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $f$ and $g$ are real and differentiable in $(a, b)$, and $g^prime(x) neq 0$ for all $x in (a, b)$, where $-infty leq a < b leq +infty$. Suppose
$$ frac{f^prime(x)}{g^prime(x)} to A mbox{ as } x to a. tag{13} $$
If
$$ f(x) to 0 mbox{ and } g(x) to 0 mbox{ as } x to a, tag{14} $$
or if
$$ g(x) to +infty mbox{ as } x to a, tag{15} $$
then
$$ frac{f(x)}{g(x)} to A mbox{ as } x to a. tag{16}$$
The analogous statement is of course also true if $x to b$, or if $g(x) to -infty$ in (15). Let us note that we now use the limit concept in the extended sense of Definition 4.33.
Here is Definition 4.33:
Let $f$ be a real function defined on $E subset mathbb{R}$. We say that
$$ f(t) to A mbox{ as } t to x, $$
where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V cap E$ is not empty, and such that $f(t) in U$ for all $t in V cap E$, $t neq x$.
And, here is Rudin's proof:
We first consider the case in which $-infty leq A < +infty$. Choose a real number $q$ such that $A < q$, and then choose $r$ such that $A < r < q$. By (13) there is a point $c in (a, b)$ such that $a < x < c$ implies
$$ frac{ f^prime(x) }{ g^prime(x) } < r. tag{17} $$
If $a < x < y < c$, then Theorem 5.9 shows that there is a point $t in (x, y)$ such that
$$ frac{ f(x)-f(y) }{ g(x)-g(y) } = frac{f^prime(t)}{g^prime(t)} < r. tag{18} $$
Suppose (14) holds. Letting $x to a$ in (18), we see that
$$ frac{f(y)}{g(y)} leq r < q qquad qquad qquad (a < y < c) tag{19} $$
Next, suppose (15) holds. Keeping $y$ fixed in (18), we can choos a point $c_1 in (a, y)$ such that $g(x) > g(y)$ and $g(x) > 0$ if $a < x < c_1$. Multiplying (18) by $left[ g(x)- g(y) right]/g(x)$, we obtain
$$ frac{ f(x) }{ g(x) } < r - r frac{ g(y) }{g(x)} + frac{f(y)}{g(x)} qquad qquad qquad (a < x < c_1). tag{20}$$
If we let $x to a$ in (20), (15) shows that there is a point $c_2 in left( a, c_1 right)$ such that
$$ frac{ f(x) }{ g(x) } < q qquad qquad qquad (a < x < c_2 ). tag{21} $$
Summing up, (19) and (21) show that for any $q$, subject only to the condition $A < q$, there is a point $c_2$ such that $f(x)/g(x) < q$ if $a < x < c_2$.
In the same manner, if $-infty < A leq +infty$, and $p$ is chosen so that $p < A$, we can find a point $c_3$ such that
$$ p < frac{ f(x) }{ g(x) } qquad qquad qquad ( a< x < c_3), tag{22} $$
and (16) follows from these two statements.
Now I have the following queries:
In (20), why has the term $f(x)/g(x)$ not been affected as we let $x to a$, although two of the three terms on the right side have gone to $0$? Don't we need any assumption about $f$ in (15)?
calculus real-analysis analysis derivatives
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
Here is Theorem 5.13 (L'Hospital's Rule) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $f$ and $g$ are real and differentiable in $(a, b)$, and $g^prime(x) neq 0$ for all $x in (a, b)$, where $-infty leq a < b leq +infty$. Suppose
$$ frac{f^prime(x)}{g^prime(x)} to A mbox{ as } x to a. tag{13} $$
If
$$ f(x) to 0 mbox{ and } g(x) to 0 mbox{ as } x to a, tag{14} $$
or if
$$ g(x) to +infty mbox{ as } x to a, tag{15} $$
then
$$ frac{f(x)}{g(x)} to A mbox{ as } x to a. tag{16}$$
The analogous statement is of course also true if $x to b$, or if $g(x) to -infty$ in (15). Let us note that we now use the limit concept in the extended sense of Definition 4.33.
Here is Definition 4.33:
Let $f$ be a real function defined on $E subset mathbb{R}$. We say that
$$ f(t) to A mbox{ as } t to x, $$
where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V cap E$ is not empty, and such that $f(t) in U$ for all $t in V cap E$, $t neq x$.
And, here is Rudin's proof:
We first consider the case in which $-infty leq A < +infty$. Choose a real number $q$ such that $A < q$, and then choose $r$ such that $A < r < q$. By (13) there is a point $c in (a, b)$ such that $a < x < c$ implies
$$ frac{ f^prime(x) }{ g^prime(x) } < r. tag{17} $$
If $a < x < y < c$, then Theorem 5.9 shows that there is a point $t in (x, y)$ such that
$$ frac{ f(x)-f(y) }{ g(x)-g(y) } = frac{f^prime(t)}{g^prime(t)} < r. tag{18} $$
Suppose (14) holds. Letting $x to a$ in (18), we see that
$$ frac{f(y)}{g(y)} leq r < q qquad qquad qquad (a < y < c) tag{19} $$
Next, suppose (15) holds. Keeping $y$ fixed in (18), we can choos a point $c_1 in (a, y)$ such that $g(x) > g(y)$ and $g(x) > 0$ if $a < x < c_1$. Multiplying (18) by $left[ g(x)- g(y) right]/g(x)$, we obtain
$$ frac{ f(x) }{ g(x) } < r - r frac{ g(y) }{g(x)} + frac{f(y)}{g(x)} qquad qquad qquad (a < x < c_1). tag{20}$$
If we let $x to a$ in (20), (15) shows that there is a point $c_2 in left( a, c_1 right)$ such that
$$ frac{ f(x) }{ g(x) } < q qquad qquad qquad (a < x < c_2 ). tag{21} $$
Summing up, (19) and (21) show that for any $q$, subject only to the condition $A < q$, there is a point $c_2$ such that $f(x)/g(x) < q$ if $a < x < c_2$.
In the same manner, if $-infty < A leq +infty$, and $p$ is chosen so that $p < A$, we can find a point $c_3$ such that
$$ p < frac{ f(x) }{ g(x) } qquad qquad qquad ( a< x < c_3), tag{22} $$
and (16) follows from these two statements.
Now I have the following queries:
In (20), why has the term $f(x)/g(x)$ not been affected as we let $x to a$, although two of the three terms on the right side have gone to $0$? Don't we need any assumption about $f$ in (15)?
calculus real-analysis analysis derivatives
calculus real-analysis analysis derivatives
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
edited May 2 '17 at 16:42
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
asked Apr 8 '17 at 19:23
Saaqib Mahmood
7,68242376
7,68242376
It is rigorous but terse. We have $f(y) -rg(y)$ fixed and $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} to 0$ as $x to a$, since $g(x) to +infty$. For $x$ sufficiently close to $a$ we have $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} < q -r$ and $f(x)/g(x) < q$.
– RRL
Apr 9 '17 at 0:09
Also, nothing about the limit of $f(x)$ or even its existence is needed to show this.
– RRL
Apr 9 '17 at 0:10
Why is Math SE voting for the closure of this question, I wonder? Is there anything about that is against the etiquette of this wonderful forum?
– Saaqib Mahmood
Apr 9 '17 at 8:33
add a comment |
It is rigorous but terse. We have $f(y) -rg(y)$ fixed and $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} to 0$ as $x to a$, since $g(x) to +infty$. For $x$ sufficiently close to $a$ we have $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} < q -r$ and $f(x)/g(x) < q$.
– RRL
Apr 9 '17 at 0:09
Also, nothing about the limit of $f(x)$ or even its existence is needed to show this.
– RRL
Apr 9 '17 at 0:10
Why is Math SE voting for the closure of this question, I wonder? Is there anything about that is against the etiquette of this wonderful forum?
– Saaqib Mahmood
Apr 9 '17 at 8:33
It is rigorous but terse. We have $f(y) -rg(y)$ fixed and $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} to 0$ as $x to a$, since $g(x) to +infty$. For $x$ sufficiently close to $a$ we have $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} < q -r$ and $f(x)/g(x) < q$.
– RRL
Apr 9 '17 at 0:09
It is rigorous but terse. We have $f(y) -rg(y)$ fixed and $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} to 0$ as $x to a$, since $g(x) to +infty$. For $x$ sufficiently close to $a$ we have $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} < q -r$ and $f(x)/g(x) < q$.
– RRL
Apr 9 '17 at 0:09
Also, nothing about the limit of $f(x)$ or even its existence is needed to show this.
– RRL
Apr 9 '17 at 0:10
Also, nothing about the limit of $f(x)$ or even its existence is needed to show this.
– RRL
Apr 9 '17 at 0:10
Why is Math SE voting for the closure of this question, I wonder? Is there anything about that is against the etiquette of this wonderful forum?
– Saaqib Mahmood
Apr 9 '17 at 8:33
Why is Math SE voting for the closure of this question, I wonder? Is there anything about that is against the etiquette of this wonderful forum?
– Saaqib Mahmood
Apr 9 '17 at 8:33
add a comment |
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It is rigorous but terse. We have $f(y) -rg(y)$ fixed and $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} to 0$ as $x to a$, since $g(x) to +infty$. For $x$ sufficiently close to $a$ we have $-r frac{g(y)}{g(x)} + frac{f(y)}{g(x)} < q -r$ and $f(x)/g(x) < q$.
– RRL
Apr 9 '17 at 0:09
Also, nothing about the limit of $f(x)$ or even its existence is needed to show this.
– RRL
Apr 9 '17 at 0:10
Why is Math SE voting for the closure of this question, I wonder? Is there anything about that is against the etiquette of this wonderful forum?
– Saaqib Mahmood
Apr 9 '17 at 8:33