Removing absolute value from argument of logarithm in $frace^x-1$ to find asymptotic
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I want to find the asymptotics of the following function when $xrightarrow0$. This is what my textbook does:
$$fracx+1e^x-1=frac(x)^1/3log(x+1)e^x-1=dots$$
I wonder why my textbook removes the absolute value? Is it because the argument of a logarithm must be positive? Any hints?
calculus
asked Aug 17 at 7:45
Cesare
56329
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up vote
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I want to find the asymptotics of the following function when $xrightarrow0$. This is what my textbook does:
$$fracx+1e^x-1=frac(x)^1/3log(x+1)e^x-1=dots$$
I wonder why my textbook removes the absolute value? Is it because the argument of a logarithm must be positive? Any hints?
calculus
asked Aug 17 at 7:45
Cesare
56329
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to find the asymptotics of the following function when $xrightarrow0$. This is what my textbook does:
$$fracx+1e^x-1=frac(x)^1/3log(x+1)e^x-1=dots$$
I wonder why my textbook removes the absolute value? Is it because the argument of a logarithm must be positive? Any hints?
calculus
asked Aug 17 at 7:45
Cesare
56329
I want to find the asymptotics of the following function when $xrightarrow0$. This is what my textbook does:
$$fracx+1e^x-1=frac(x)^1/3log(x+1)e^x-1=dots$$
I wonder why my textbook removes the absolute value? Is it because the argument of a logarithm must be positive? Any hints?
calculus
asked Aug 17 at 7:45
Cesare
56329
asked Aug 17 at 7:45
Cesare
56329
asked Aug 17 at 7:45
Cesare
56329
asked Aug 17 at 7:45
Cesare
56329
56329
add a comment |Â
add a comment |Â
2 Answers
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4
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accepted
For $x$ "near" $0$ we have $x+1$ "near" $1$, hence in the considerations concerning $x to 0$, we can assume that $x+1>0$.
answered Aug 17 at 7:51
Fred
38.2k1238
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up vote
2
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The logarithm is defined only when it's argument is positive. When you have an absolute value as the argument of the logarithm the latter is defined $forall x$ and is positive as $x$ approaches zero because you have $x+1$ as the argument.
If there wasn't that $+1$ and your function was something like this $$f(x)=fracxe^x-1$$ then you have to split the limit as$x$ goes to zero in two parts, mainly $$limlimits_xrightarrow 0^+fracx^1over3log(x)e^x-1\limlimits_xrightarrow 0^-fracx^1over3log(-x)e^x-1$$
answered Aug 17 at 7:53
Davide Morgante
2,336322
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For $x$ "near" $0$ we have $x+1$ "near" $1$, hence in the considerations concerning $x to 0$, we can assume that $x+1>0$.
answered Aug 17 at 7:51
Fred
38.2k1238
add a comment |Â
up vote
4
down vote
accepted
For $x$ "near" $0$ we have $x+1$ "near" $1$, hence in the considerations concerning $x to 0$, we can assume that $x+1>0$.
answered Aug 17 at 7:51
Fred
38.2k1238
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For $x$ "near" $0$ we have $x+1$ "near" $1$, hence in the considerations concerning $x to 0$, we can assume that $x+1>0$.
answered Aug 17 at 7:51
Fred
38.2k1238
For $x$ "near" $0$ we have $x+1$ "near" $1$, hence in the considerations concerning $x to 0$, we can assume that $x+1>0$.
answered Aug 17 at 7:51
Fred
38.2k1238
answered Aug 17 at 7:51
Fred
38.2k1238
answered Aug 17 at 7:51
Fred
38.2k1238
answered Aug 17 at 7:51
Fred
38.2k1238
38.2k1238
add a comment |Â
add a comment |Â
up vote
2
down vote
The logarithm is defined only when it's argument is positive. When you have an absolute value as the argument of the logarithm the latter is defined $forall x$ and is positive as $x$ approaches zero because you have $x+1$ as the argument.
If there wasn't that $+1$ and your function was something like this $$f(x)=fracxe^x-1$$ then you have to split the limit as$x$ goes to zero in two parts, mainly $$limlimits_xrightarrow 0^+fracx^1over3log(x)e^x-1\limlimits_xrightarrow 0^-fracx^1over3log(-x)e^x-1$$
answered Aug 17 at 7:53
Davide Morgante
2,336322
add a comment |Â
up vote
2
down vote
The logarithm is defined only when it's argument is positive. When you have an absolute value as the argument of the logarithm the latter is defined $forall x$ and is positive as $x$ approaches zero because you have $x+1$ as the argument.
If there wasn't that $+1$ and your function was something like this $$f(x)=fracxe^x-1$$ then you have to split the limit as$x$ goes to zero in two parts, mainly $$limlimits_xrightarrow 0^+fracx^1over3log(x)e^x-1\limlimits_xrightarrow 0^-fracx^1over3log(-x)e^x-1$$
answered Aug 17 at 7:53
Davide Morgante
2,336322
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The logarithm is defined only when it's argument is positive. When you have an absolute value as the argument of the logarithm the latter is defined $forall x$ and is positive as $x$ approaches zero because you have $x+1$ as the argument.
If there wasn't that $+1$ and your function was something like this $$f(x)=fracxe^x-1$$ then you have to split the limit as$x$ goes to zero in two parts, mainly $$limlimits_xrightarrow 0^+fracx^1over3log(x)e^x-1\limlimits_xrightarrow 0^-fracx^1over3log(-x)e^x-1$$
answered Aug 17 at 7:53
Davide Morgante
2,336322
The logarithm is defined only when it's argument is positive. When you have an absolute value as the argument of the logarithm the latter is defined $forall x$ and is positive as $x$ approaches zero because you have $x+1$ as the argument.
If there wasn't that $+1$ and your function was something like this $$f(x)=fracxe^x-1$$ then you have to split the limit as$x$ goes to zero in two parts, mainly $$limlimits_xrightarrow 0^+fracx^1over3log(x)e^x-1\limlimits_xrightarrow 0^-fracx^1over3log(-x)e^x-1$$
answered Aug 17 at 7:53
Davide Morgante
2,336322
answered Aug 17 at 7:53
Davide Morgante
2,336322
answered Aug 17 at 7:53
Davide Morgante
2,336322
answered Aug 17 at 7:53
Davide Morgante
2,336322
2,336322
add a comment |Â
add a comment |Â
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