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Rudin's Proof on L'hospital's Rule
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I have a little problem understanding a point of Rudin's proof on L'hopital's Rule. Here is the statement of the L'hospital's Rule:
L'hospital's Rule Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)neq 0$ for all $xin (a,b)$, where $-infty leq a<bleq infty$. Suppose $fracf'(x)g'(x)rightarrow A$ as $xrightarrow a$. If $f(x)rightarrow 0$ and $g(x)rightarrow 0$ as $xrightarrow a$, then $fracf(x)g(x) rightarrow A$ as $xrightarrow a$.
Rudin uses the generalized mean value theorem (GMVT):
GMVT If f and g are continuous real functions on [a, b] which are differentiable in (a, b), then there is a point $xin (a,b)$ at which $[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)$.
Here is Rudin's proof on L'hospital's Rule:
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 $A<r<q$. Then, there is a point $cin (a,b)$ such that $a<x<c$ implies $fracf'(x)g'(x)<r$. If $a<x<y<c$, then GMVT shows that there is a point $tin (x, y)$ such that $fracf(x)-f(y)g(x)-g(y)=fracf'(t)g'(t)<r$.
For the last line of the above paragraph, I am a little confused on how can we make sure that $g(x)neq g(y)$? Even though $g'(x)neq 0$ for all $xin (a,b)$, we still may have $g(x)=g(y)$ for some $x$ and $y$ in $(a,c)$. If $g(x)=g(y)$, we have a fraction over $0$ which can go to infinity, and thus cannot be less than $rin R$.
real-analysis proof-explanation
asked Jul 30 at 1:03
James Wang
876
add a comment |Â
up vote
0
down vote
favorite
I have a little problem understanding a point of Rudin's proof on L'hopital's Rule. Here is the statement of the L'hospital's Rule:
L'hospital's Rule Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)neq 0$ for all $xin (a,b)$, where $-infty leq a<bleq infty$. Suppose $fracf'(x)g'(x)rightarrow A$ as $xrightarrow a$. If $f(x)rightarrow 0$ and $g(x)rightarrow 0$ as $xrightarrow a$, then $fracf(x)g(x) rightarrow A$ as $xrightarrow a$.
Rudin uses the generalized mean value theorem (GMVT):
GMVT If f and g are continuous real functions on [a, b] which are differentiable in (a, b), then there is a point $xin (a,b)$ at which $[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)$.
Here is Rudin's proof on L'hospital's Rule:
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 $A<r<q$. Then, there is a point $cin (a,b)$ such that $a<x<c$ implies $fracf'(x)g'(x)<r$. If $a<x<y<c$, then GMVT shows that there is a point $tin (x, y)$ such that $fracf(x)-f(y)g(x)-g(y)=fracf'(t)g'(t)<r$.
For the last line of the above paragraph, I am a little confused on how can we make sure that $g(x)neq g(y)$? Even though $g'(x)neq 0$ for all $xin (a,b)$, we still may have $g(x)=g(y)$ for some $x$ and $y$ in $(a,c)$. If $g(x)=g(y)$, we have a fraction over $0$ which can go to infinity, and thus cannot be less than $rin R$.
real-analysis proof-explanation
asked Jul 30 at 1:03
James Wang
876
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a little problem understanding a point of Rudin's proof on L'hopital's Rule. Here is the statement of the L'hospital's Rule:
L'hospital's Rule Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)neq 0$ for all $xin (a,b)$, where $-infty leq a<bleq infty$. Suppose $fracf'(x)g'(x)rightarrow A$ as $xrightarrow a$. If $f(x)rightarrow 0$ and $g(x)rightarrow 0$ as $xrightarrow a$, then $fracf(x)g(x) rightarrow A$ as $xrightarrow a$.
Rudin uses the generalized mean value theorem (GMVT):
GMVT If f and g are continuous real functions on [a, b] which are differentiable in (a, b), then there is a point $xin (a,b)$ at which $[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)$.
Here is Rudin's proof on L'hospital's Rule:
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 $A<r<q$. Then, there is a point $cin (a,b)$ such that $a<x<c$ implies $fracf'(x)g'(x)<r$. If $a<x<y<c$, then GMVT shows that there is a point $tin (x, y)$ such that $fracf(x)-f(y)g(x)-g(y)=fracf'(t)g'(t)<r$.
For the last line of the above paragraph, I am a little confused on how can we make sure that $g(x)neq g(y)$? Even though $g'(x)neq 0$ for all $xin (a,b)$, we still may have $g(x)=g(y)$ for some $x$ and $y$ in $(a,c)$. If $g(x)=g(y)$, we have a fraction over $0$ which can go to infinity, and thus cannot be less than $rin R$.
real-analysis proof-explanation
asked Jul 30 at 1:03
James Wang
876
I have a little problem understanding a point of Rudin's proof on L'hopital's Rule. Here is the statement of the L'hospital's Rule:
L'hospital's Rule Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)neq 0$ for all $xin (a,b)$, where $-infty leq a<bleq infty$. Suppose $fracf'(x)g'(x)rightarrow A$ as $xrightarrow a$. If $f(x)rightarrow 0$ and $g(x)rightarrow 0$ as $xrightarrow a$, then $fracf(x)g(x) rightarrow A$ as $xrightarrow a$.
Rudin uses the generalized mean value theorem (GMVT):
GMVT If f and g are continuous real functions on [a, b] which are differentiable in (a, b), then there is a point $xin (a,b)$ at which $[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)$.
Here is Rudin's proof on L'hospital's Rule:
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 $A<r<q$. Then, there is a point $cin (a,b)$ such that $a<x<c$ implies $fracf'(x)g'(x)<r$. If $a<x<y<c$, then GMVT shows that there is a point $tin (x, y)$ such that $fracf(x)-f(y)g(x)-g(y)=fracf'(t)g'(t)<r$.
For the last line of the above paragraph, I am a little confused on how can we make sure that $g(x)neq g(y)$? Even though $g'(x)neq 0$ for all $xin (a,b)$, we still may have $g(x)=g(y)$ for some $x$ and $y$ in $(a,c)$. If $g(x)=g(y)$, we have a fraction over $0$ which can go to infinity, and thus cannot be less than $rin R$.
real-analysis proof-explanation
asked Jul 30 at 1:03
James Wang
876
asked Jul 30 at 1:03
James Wang
876
asked Jul 30 at 1:03
James Wang
876
asked Jul 30 at 1:03
James Wang
876
876
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
By Rolle's Theorem, if $g(x)=g(y)$ for some $x <y$, then, there exists some $c in (x,y)$ such that $g'(c)=0$.
answered Jul 30 at 1:07
N. S.
97.7k5105197
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By Rolle's Theorem, if $g(x)=g(y)$ for some $x <y$, then, there exists some $c in (x,y)$ such that $g'(c)=0$.
answered Jul 30 at 1:07
N. S.
97.7k5105197
add a comment |Â
up vote
1
down vote
accepted
By Rolle's Theorem, if $g(x)=g(y)$ for some $x <y$, then, there exists some $c in (x,y)$ such that $g'(c)=0$.
answered Jul 30 at 1:07
N. S.
97.7k5105197
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By Rolle's Theorem, if $g(x)=g(y)$ for some $x <y$, then, there exists some $c in (x,y)$ such that $g'(c)=0$.
answered Jul 30 at 1:07
N. S.
97.7k5105197
By Rolle's Theorem, if $g(x)=g(y)$ for some $x <y$, then, there exists some $c in (x,y)$ such that $g'(c)=0$.
answered Jul 30 at 1:07
N. S.
97.7k5105197
answered Jul 30 at 1:07
N. S.
97.7k5105197
answered Jul 30 at 1:07
N. S.
97.7k5105197
answered Jul 30 at 1:07
N. S.
97.7k5105197
97.7k5105197
add a comment |Â
add a comment |Â
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