Mixing
Order of $f(x)=frac{3}{-2ln|x|-2ln|x+3|}$ with respect to $x$
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I have $f(x)=frac{3}{-2ln|x|-2ln|x+3|}$ and have to determine the order of this infinitesimal with respect to $x$ for $xto0$.Then it doesn't matter if we are approaching from left or right because $-2ln|x|to+infty$ and : $$lim_{xto0}frac{3}{-2ln|x|x^alpha}$$ But it doesn't seem like for any $alpha$ we can have limit approaching a real number.How can order be determined in that case?
calculus limits infinitesimals
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
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show 1 more comment
$begingroup$
I have $f(x)=frac{3}{-2ln|x|-2ln|x+3|}$ and have to determine the order of this infinitesimal with respect to $x$ for $xto0$.Then it doesn't matter if we are approaching from left or right because $-2ln|x|to+infty$ and : $$lim_{xto0}frac{3}{-2ln|x|x^alpha}$$ But it doesn't seem like for any $alpha$ we can have limit approaching a real number.How can order be determined in that case?
calculus limits infinitesimals
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
$endgroup$
$begingroup$
What if $alpha <0$?
$endgroup$
– Botond
Jan 26 at 23:01
$begingroup$
If $alphalt0$ we would have $x^alpha$ in numerator with $alphagt0$.But still I don't get by what kind of logic we can have a real valued limit for function.
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– Turan Nasibli
Jan 26 at 23:09
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Can you calculate the limit, for example, when $alpha=-1$?
$endgroup$
– Botond
Jan 27 at 0:08
$begingroup$
@Botond Yes,the limit is equal to $0$
$endgroup$
– Turan Nasibli
Jan 27 at 1:36
$begingroup$
Isn't that a real valued limit?
$endgroup$
– Botond
Jan 27 at 8:59
|
show 1 more comment
$begingroup$
I have $f(x)=frac{3}{-2ln|x|-2ln|x+3|}$ and have to determine the order of this infinitesimal with respect to $x$ for $xto0$.Then it doesn't matter if we are approaching from left or right because $-2ln|x|to+infty$ and : $$lim_{xto0}frac{3}{-2ln|x|x^alpha}$$ But it doesn't seem like for any $alpha$ we can have limit approaching a real number.How can order be determined in that case?
calculus limits infinitesimals
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
$endgroup$
I have $f(x)=frac{3}{-2ln|x|-2ln|x+3|}$ and have to determine the order of this infinitesimal with respect to $x$ for $xto0$.Then it doesn't matter if we are approaching from left or right because $-2ln|x|to+infty$ and : $$lim_{xto0}frac{3}{-2ln|x|x^alpha}$$ But it doesn't seem like for any $alpha$ we can have limit approaching a real number.How can order be determined in that case?
calculus limits infinitesimals
calculus limits infinitesimals
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
edited Jan 26 at 22:54
Turan Nasibli
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
asked Jan 26 at 22:37
Turan NasibliTuran Nasibli
846
846
$begingroup$
What if $alpha <0$?
$endgroup$
– Botond
Jan 26 at 23:01
$begingroup$
If $alphalt0$ we would have $x^alpha$ in numerator with $alphagt0$.But still I don't get by what kind of logic we can have a real valued limit for function.
$endgroup$
– Turan Nasibli
Jan 26 at 23:09
$begingroup$
Can you calculate the limit, for example, when $alpha=-1$?
$endgroup$
– Botond
Jan 27 at 0:08
$begingroup$
@Botond Yes,the limit is equal to $0$
$endgroup$
– Turan Nasibli
Jan 27 at 1:36
$begingroup$
Isn't that a real valued limit?
$endgroup$
– Botond
Jan 27 at 8:59
|
show 1 more comment
$begingroup$
What if $alpha <0$?
$endgroup$
– Botond
Jan 26 at 23:01
$begingroup$
If $alphalt0$ we would have $x^alpha$ in numerator with $alphagt0$.But still I don't get by what kind of logic we can have a real valued limit for function.
$endgroup$
– Turan Nasibli
Jan 26 at 23:09
$begingroup$
Can you calculate the limit, for example, when $alpha=-1$?
$endgroup$
– Botond
Jan 27 at 0:08
$begingroup$
@Botond Yes,the limit is equal to $0$
$endgroup$
– Turan Nasibli
Jan 27 at 1:36
$begingroup$
Isn't that a real valued limit?
$endgroup$
– Botond
Jan 27 at 8:59
$begingroup$
What if $alpha <0$?
$endgroup$
– Botond
Jan 26 at 23:01
$begingroup$
What if $alpha <0$?
$endgroup$
– Botond
Jan 26 at 23:01
$begingroup$
If $alphalt0$ we would have $x^alpha$ in numerator with $alphagt0$.But still I don't get by what kind of logic we can have a real valued limit for function.
$endgroup$
– Turan Nasibli
Jan 26 at 23:09
$begingroup$
If $alphalt0$ we would have $x^alpha$ in numerator with $alphagt0$.But still I don't get by what kind of logic we can have a real valued limit for function.
$endgroup$
– Turan Nasibli
Jan 26 at 23:09
$begingroup$
Can you calculate the limit, for example, when $alpha=-1$?
$endgroup$
– Botond
Jan 27 at 0:08
$begingroup$
Can you calculate the limit, for example, when $alpha=-1$?
$endgroup$
– Botond
Jan 27 at 0:08
$begingroup$
@Botond Yes,the limit is equal to $0$
$endgroup$
– Turan Nasibli
Jan 27 at 1:36
$begingroup$
@Botond Yes,the limit is equal to $0$
$endgroup$
– Turan Nasibli
Jan 27 at 1:36
$begingroup$
Isn't that a real valued limit?
$endgroup$
– Botond
Jan 27 at 8:59
$begingroup$
Isn't that a real valued limit?
$endgroup$
– Botond
Jan 27 at 8:59
|
show 1 more comment
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$begingroup$
What if $alpha <0$?
$endgroup$
– Botond
Jan 26 at 23:01
$begingroup$
If $alphalt0$ we would have $x^alpha$ in numerator with $alphagt0$.But still I don't get by what kind of logic we can have a real valued limit for function.
$endgroup$
– Turan Nasibli
Jan 26 at 23:09
$begingroup$
Can you calculate the limit, for example, when $alpha=-1$?
$endgroup$
– Botond
Jan 27 at 0:08
$begingroup$
@Botond Yes,the limit is equal to $0$
$endgroup$
– Turan Nasibli
Jan 27 at 1:36
$begingroup$
Isn't that a real valued limit?
$endgroup$
– Botond
Jan 27 at 8:59