How to be sure here that we are not dividing by zero?
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This is from baby Rudin Ch 5, exer 10:
Suppose $f$ and $g$ are complex differentiable functions on $(0,1)$, $f(x)to0, g(x)to0, f'(x)to A, g'(x)to B$ as $xto0$, where $A$ and $B$ are complex numbers, $Bne0$. Prove that $$limlimits_{xto0}frac{f(x)}{g(x)}=frac{A}{B}.$$
What I can't get is how do we know that $g(x)ne0$ in an nbd of $0$.
real-analysis limits derivatives
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
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add a comment |
$begingroup$
This is from baby Rudin Ch 5, exer 10:
Suppose $f$ and $g$ are complex differentiable functions on $(0,1)$, $f(x)to0, g(x)to0, f'(x)to A, g'(x)to B$ as $xto0$, where $A$ and $B$ are complex numbers, $Bne0$. Prove that $$limlimits_{xto0}frac{f(x)}{g(x)}=frac{A}{B}.$$
What I can't get is how do we know that $g(x)ne0$ in an nbd of $0$.
real-analysis limits derivatives
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
$endgroup$
2
$begingroup$
The Taylor series of $g$ around $0$ is $g(x) = 0 + Bx + O(x^2)$. In a neighbourhood of $0$, it has to behave like $Bx$.
$endgroup$
– Luke
Nov 27 '18 at 10:48
add a comment |
$begingroup$
This is from baby Rudin Ch 5, exer 10:
Suppose $f$ and $g$ are complex differentiable functions on $(0,1)$, $f(x)to0, g(x)to0, f'(x)to A, g'(x)to B$ as $xto0$, where $A$ and $B$ are complex numbers, $Bne0$. Prove that $$limlimits_{xto0}frac{f(x)}{g(x)}=frac{A}{B}.$$
What I can't get is how do we know that $g(x)ne0$ in an nbd of $0$.
real-analysis limits derivatives
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
$endgroup$
This is from baby Rudin Ch 5, exer 10:
Suppose $f$ and $g$ are complex differentiable functions on $(0,1)$, $f(x)to0, g(x)to0, f'(x)to A, g'(x)to B$ as $xto0$, where $A$ and $B$ are complex numbers, $Bne0$. Prove that $$limlimits_{xto0}frac{f(x)}{g(x)}=frac{A}{B}.$$
What I can't get is how do we know that $g(x)ne0$ in an nbd of $0$.
real-analysis limits derivatives
real-analysis limits derivatives
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
asked Nov 27 '18 at 10:41
SilentSilent
2,77532150
2,77532150
2
$begingroup$
The Taylor series of $g$ around $0$ is $g(x) = 0 + Bx + O(x^2)$. In a neighbourhood of $0$, it has to behave like $Bx$.
$endgroup$
– Luke
Nov 27 '18 at 10:48
add a comment |
2
$begingroup$
The Taylor series of $g$ around $0$ is $g(x) = 0 + Bx + O(x^2)$. In a neighbourhood of $0$, it has to behave like $Bx$.
$endgroup$
– Luke
Nov 27 '18 at 10:48
2
2
$begingroup$
The Taylor series of $g$ around $0$ is $g(x) = 0 + Bx + O(x^2)$. In a neighbourhood of $0$, it has to behave like $Bx$.
$endgroup$
– Luke
Nov 27 '18 at 10:48
$begingroup$
The Taylor series of $g$ around $0$ is $g(x) = 0 + Bx + O(x^2)$. In a neighbourhood of $0$, it has to behave like $Bx$.
$endgroup$
– Luke
Nov 27 '18 at 10:48
add a comment |
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$begingroup$
The Taylor series of $g$ around $0$ is $g(x) = 0 + Bx + O(x^2)$. In a neighbourhood of $0$, it has to behave like $Bx$.
$endgroup$
– Luke
Nov 27 '18 at 10:48