Is this “inverse the limit†process right?
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Suppose you have two (nice enough) functions $f$ and $g$ and a constant $lambda$ such that $$lim_xtoinftyfracf(x)g(x)=lambda$$
Is it true that $$lim_xtoinftyfracf^-1(x)g^-1(x/lambda)=1$$
The "reasoning" goes like this: $$fracf(x)g(x)approxlambda$$ $$f(x)approxlambda g(x)$$ $$xapprox f^-1(lambda g(x))$$ $$g^-1(x/lambda)approx f^-1(x)$$ $$fracf^-1(x)g^-1(x/lambda)approx 1$$
all of this supposing there is no problem in $xmapsto f^-1(x)$ and $xmapsto g^-1(x/lambda)$
I think assuming "nice enough" (continuity, inverse, ...) and being a bit more precise like $$f(x)=lambda g(x)+o(g(x))$$ will prove the statement.
What I'm more interested in is under what conditions does it remain true in discrete variable (and non-existing inverse function for $f$ or $g$ or both)
Thanks!
real-analysis limits
asked Aug 18 at 23:23
Pedro
474212
 |Â
show 1 more comment
up vote
10
down vote
favorite
Suppose you have two (nice enough) functions $f$ and $g$ and a constant $lambda$ such that $$lim_xtoinftyfracf(x)g(x)=lambda$$
Is it true that $$lim_xtoinftyfracf^-1(x)g^-1(x/lambda)=1$$
The "reasoning" goes like this: $$fracf(x)g(x)approxlambda$$ $$f(x)approxlambda g(x)$$ $$xapprox f^-1(lambda g(x))$$ $$g^-1(x/lambda)approx f^-1(x)$$ $$fracf^-1(x)g^-1(x/lambda)approx 1$$
all of this supposing there is no problem in $xmapsto f^-1(x)$ and $xmapsto g^-1(x/lambda)$
I think assuming "nice enough" (continuity, inverse, ...) and being a bit more precise like $$f(x)=lambda g(x)+o(g(x))$$ will prove the statement.
What I'm more interested in is under what conditions does it remain true in discrete variable (and non-existing inverse function for $f$ or $g$ or both)
Thanks!
real-analysis limits
asked Aug 18 at 23:23
Pedro
474212
Why do you have $n$ in the limit?
– Thomas Andrews
Aug 18 at 23:28
6
Where does $g^-1(x/lambda)approx f^-1(x)$ come from?
– Jack M
Aug 19 at 0:36
@ThomasAndrews thanks! i wanted to write $x$
– Pedro
Aug 19 at 1:05
@JackM from setting $xmapsto g^-1(x/lambda )$
– Pedro
Aug 19 at 1:06
@Pedro I don't understand what that means. We have $g(x)approx fracf(x)lambda$, but not $g(x)approxfrac xlambda$
– Jack M
Aug 19 at 10:35
 |Â
show 1 more comment
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Suppose you have two (nice enough) functions $f$ and $g$ and a constant $lambda$ such that $$lim_xtoinftyfracf(x)g(x)=lambda$$
Is it true that $$lim_xtoinftyfracf^-1(x)g^-1(x/lambda)=1$$
The "reasoning" goes like this: $$fracf(x)g(x)approxlambda$$ $$f(x)approxlambda g(x)$$ $$xapprox f^-1(lambda g(x))$$ $$g^-1(x/lambda)approx f^-1(x)$$ $$fracf^-1(x)g^-1(x/lambda)approx 1$$
all of this supposing there is no problem in $xmapsto f^-1(x)$ and $xmapsto g^-1(x/lambda)$
I think assuming "nice enough" (continuity, inverse, ...) and being a bit more precise like $$f(x)=lambda g(x)+o(g(x))$$ will prove the statement.
What I'm more interested in is under what conditions does it remain true in discrete variable (and non-existing inverse function for $f$ or $g$ or both)
Thanks!
real-analysis limits
asked Aug 18 at 23:23
Pedro
474212
Suppose you have two (nice enough) functions $f$ and $g$ and a constant $lambda$ such that $$lim_xtoinftyfracf(x)g(x)=lambda$$
Is it true that $$lim_xtoinftyfracf^-1(x)g^-1(x/lambda)=1$$
The "reasoning" goes like this: $$fracf(x)g(x)approxlambda$$ $$f(x)approxlambda g(x)$$ $$xapprox f^-1(lambda g(x))$$ $$g^-1(x/lambda)approx f^-1(x)$$ $$fracf^-1(x)g^-1(x/lambda)approx 1$$
all of this supposing there is no problem in $xmapsto f^-1(x)$ and $xmapsto g^-1(x/lambda)$
I think assuming "nice enough" (continuity, inverse, ...) and being a bit more precise like $$f(x)=lambda g(x)+o(g(x))$$ will prove the statement.
What I'm more interested in is under what conditions does it remain true in discrete variable (and non-existing inverse function for $f$ or $g$ or both)
Thanks!
real-analysis limits
real-analysis limits
asked Aug 18 at 23:23
Pedro
474212
asked Aug 18 at 23:23
Pedro
474212
edited Aug 19 at 1:04
asked Aug 18 at 23:23
Pedro
474212
asked Aug 18 at 23:23
Pedro
474212
asked Aug 18 at 23:23
Pedro
474212
474212
Why do you have $n$ in the limit?
– Thomas Andrews
Aug 18 at 23:28
6
Where does $g^-1(x/lambda)approx f^-1(x)$ come from?
– Jack M
Aug 19 at 0:36
@ThomasAndrews thanks! i wanted to write $x$
– Pedro
Aug 19 at 1:05
@JackM from setting $xmapsto g^-1(x/lambda )$
– Pedro
Aug 19 at 1:06
@Pedro I don't understand what that means. We have $g(x)approx fracf(x)lambda$, but not $g(x)approxfrac xlambda$
– Jack M
Aug 19 at 10:35
 |Â
show 1 more comment
Why do you have $n$ in the limit?
– Thomas Andrews
Aug 18 at 23:28
6
Where does $g^-1(x/lambda)approx f^-1(x)$ come from?
– Jack M
Aug 19 at 0:36
@ThomasAndrews thanks! i wanted to write $x$
– Pedro
Aug 19 at 1:05
@JackM from setting $xmapsto g^-1(x/lambda )$
– Pedro
Aug 19 at 1:06
@Pedro I don't understand what that means. We have $g(x)approx fracf(x)lambda$, but not $g(x)approxfrac xlambda$
– Jack M
Aug 19 at 10:35
Why do you have $n$ in the limit?
– Thomas Andrews
Aug 18 at 23:28
Why do you have $n$ in the limit?
– Thomas Andrews
Aug 18 at 23:28
6
6
Where does $g^-1(x/lambda)approx f^-1(x)$ come from?
– Jack M
Aug 19 at 0:36
Where does $g^-1(x/lambda)approx f^-1(x)$ come from?
– Jack M
Aug 19 at 0:36
@ThomasAndrews thanks! i wanted to write $x$
– Pedro
Aug 19 at 1:05
@ThomasAndrews thanks! i wanted to write $x$
– Pedro
Aug 19 at 1:05
@JackM from setting $xmapsto g^-1(x/lambda )$
– Pedro
Aug 19 at 1:06
@JackM from setting $xmapsto g^-1(x/lambda )$
– Pedro
Aug 19 at 1:06
@Pedro I don't understand what that means. We have $g(x)approx fracf(x)lambda$, but not $g(x)approxfrac xlambda$
– Jack M
Aug 19 at 10:35
@Pedro I don't understand what that means. We have $g(x)approx fracf(x)lambda$, but not $g(x)approxfrac xlambda$
– Jack M
Aug 19 at 10:35
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
17
down vote
accepted
No. If you take $f(x) = log(x)$ and $g(x) = log(x)-1$ and $lambda = 1$ then :
$$ undersetxrightarrowinftylimfracf(x)g(x) = 1 $$
But :
$$ fracf^-1(x)g^-1(x) = frace^xe^x+1 = e^-1 neq 1$$
answered Aug 18 at 23:35
tmaths
1,336113
Nice counterexample! Thanks
– Pedro
Aug 19 at 1:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
No. If you take $f(x) = log(x)$ and $g(x) = log(x)-1$ and $lambda = 1$ then :
$$ undersetxrightarrowinftylimfracf(x)g(x) = 1 $$
But :
$$ fracf^-1(x)g^-1(x) = frace^xe^x+1 = e^-1 neq 1$$
answered Aug 18 at 23:35
tmaths
1,336113
Nice counterexample! Thanks
– Pedro
Aug 19 at 1:10
add a comment |Â
up vote
17
down vote
accepted
No. If you take $f(x) = log(x)$ and $g(x) = log(x)-1$ and $lambda = 1$ then :
$$ undersetxrightarrowinftylimfracf(x)g(x) = 1 $$
But :
$$ fracf^-1(x)g^-1(x) = frace^xe^x+1 = e^-1 neq 1$$
answered Aug 18 at 23:35
tmaths
1,336113
Nice counterexample! Thanks
– Pedro
Aug 19 at 1:10
add a comment |Â
up vote
17
down vote
accepted
up vote
17
down vote
accepted
No. If you take $f(x) = log(x)$ and $g(x) = log(x)-1$ and $lambda = 1$ then :
$$ undersetxrightarrowinftylimfracf(x)g(x) = 1 $$
But :
$$ fracf^-1(x)g^-1(x) = frace^xe^x+1 = e^-1 neq 1$$
answered Aug 18 at 23:35
tmaths
1,336113
No. If you take $f(x) = log(x)$ and $g(x) = log(x)-1$ and $lambda = 1$ then :
$$ undersetxrightarrowinftylimfracf(x)g(x) = 1 $$
But :
$$ fracf^-1(x)g^-1(x) = frace^xe^x+1 = e^-1 neq 1$$
answered Aug 18 at 23:35
tmaths
1,336113
answered Aug 18 at 23:35
tmaths
1,336113
answered Aug 18 at 23:35
tmaths
1,336113
answered Aug 18 at 23:35
tmaths
1,336113
1,336113
Nice counterexample! Thanks
– Pedro
Aug 19 at 1:10
add a comment |Â
Nice counterexample! Thanks
– Pedro
Aug 19 at 1:10
Nice counterexample! Thanks
– Pedro
Aug 19 at 1:10
Nice counterexample! Thanks
– Pedro
Aug 19 at 1:10
add a comment |Â
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Why do you have $n$ in the limit?
– Thomas Andrews
Aug 18 at 23:28
6
Where does $g^-1(x/lambda)approx f^-1(x)$ come from?
– Jack M
Aug 19 at 0:36
@ThomasAndrews thanks! i wanted to write $x$
– Pedro
Aug 19 at 1:05
@JackM from setting $xmapsto g^-1(x/lambda )$
– Pedro
Aug 19 at 1:06
@Pedro I don't understand what that means. We have $g(x)approx fracf(x)lambda$, but not $g(x)approxfrac xlambda$
– Jack M
Aug 19 at 10:35