Justify $e^2xlogxsim (x-1)$
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How do I justify that for $xrightarrow1$
$$e^2xlogxsim (x-1)$$
I know that if $xrightarrow 0$, then $logxsim(x-1)$ because $log(1+x)sim x$ for $xrightarrow0$. This limit does not approach $0$, though, but 1. Any hints?
This is what my textbook does:
$$frace^2xlogx(x-1)^1/3simfrac(x-1)(x-1)^1/3$$
calculus asymptotics
asked Aug 17 at 6:55
Cesare
56329
add a comment |Â
up vote
0
down vote
favorite
How do I justify that for $xrightarrow1$
$$e^2xlogxsim (x-1)$$
I know that if $xrightarrow 0$, then $logxsim(x-1)$ because $log(1+x)sim x$ for $xrightarrow0$. This limit does not approach $0$, though, but 1. Any hints?
This is what my textbook does:
$$frace^2xlogx(x-1)^1/3simfrac(x-1)(x-1)^1/3$$
calculus asymptotics
asked Aug 17 at 6:55
Cesare
56329
That is a mistake. If by textbook you mean lecture notes, you should point this out to the professor who wrote it. It will be appreciated. (If it is a proper book, you can also write an e-mail to the author.)
– A. Pongrácz
Aug 17 at 8:23
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I justify that for $xrightarrow1$
$$e^2xlogxsim (x-1)$$
I know that if $xrightarrow 0$, then $logxsim(x-1)$ because $log(1+x)sim x$ for $xrightarrow0$. This limit does not approach $0$, though, but 1. Any hints?
This is what my textbook does:
$$frace^2xlogx(x-1)^1/3simfrac(x-1)(x-1)^1/3$$
calculus asymptotics
asked Aug 17 at 6:55
Cesare
56329
How do I justify that for $xrightarrow1$
$$e^2xlogxsim (x-1)$$
I know that if $xrightarrow 0$, then $logxsim(x-1)$ because $log(1+x)sim x$ for $xrightarrow0$. This limit does not approach $0$, though, but 1. Any hints?
This is what my textbook does:
$$frace^2xlogx(x-1)^1/3simfrac(x-1)(x-1)^1/3$$
calculus asymptotics
asked Aug 17 at 6:55
Cesare
56329
edited Aug 17 at 7:02
asked Aug 17 at 6:55
Cesare
56329
asked Aug 17 at 6:55
Cesare
56329
asked Aug 17 at 6:55
Cesare
56329
56329
That is a mistake. If by textbook you mean lecture notes, you should point this out to the professor who wrote it. It will be appreciated. (If it is a proper book, you can also write an e-mail to the author.)
– A. Pongrácz
Aug 17 at 8:23
add a comment |Â
That is a mistake. If by textbook you mean lecture notes, you should point this out to the professor who wrote it. It will be appreciated. (If it is a proper book, you can also write an e-mail to the author.)
– A. Pongrácz
Aug 17 at 8:23
That is a mistake. If by textbook you mean lecture notes, you should point this out to the professor who wrote it. It will be appreciated. (If it is a proper book, you can also write an e-mail to the author.)
– A. Pongrácz
Aug 17 at 8:23
That is a mistake. If by textbook you mean lecture notes, you should point this out to the professor who wrote it. It will be appreciated. (If it is a proper book, you can also write an e-mail to the author.)
– A. Pongrácz
Aug 17 at 8:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
As $xrightarrow 1$, we have $log xsim x-1$, and $e^2xsim e^2$, so $e^2xlog xsim e^2(x-1)$.
answered Aug 17 at 7:00
A. Pongrácz
3,877625
add a comment |Â
up vote
2
down vote
By L'Hospital we have
$lim_x to 1frace^2xlogxx-1= lim_x to 1(2e^2x log x+e^2xfrac1x)=e^2$.
answered Aug 17 at 7:13
Fred
38.2k1238
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
As $xrightarrow 1$, we have $log xsim x-1$, and $e^2xsim e^2$, so $e^2xlog xsim e^2(x-1)$.
answered Aug 17 at 7:00
A. Pongrácz
3,877625
add a comment |Â
up vote
4
down vote
accepted
As $xrightarrow 1$, we have $log xsim x-1$, and $e^2xsim e^2$, so $e^2xlog xsim e^2(x-1)$.
answered Aug 17 at 7:00
A. Pongrácz
3,877625
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
As $xrightarrow 1$, we have $log xsim x-1$, and $e^2xsim e^2$, so $e^2xlog xsim e^2(x-1)$.
answered Aug 17 at 7:00
A. Pongrácz
3,877625
As $xrightarrow 1$, we have $log xsim x-1$, and $e^2xsim e^2$, so $e^2xlog xsim e^2(x-1)$.
answered Aug 17 at 7:00
A. Pongrácz
3,877625
answered Aug 17 at 7:00
A. Pongrácz
3,877625
answered Aug 17 at 7:00
A. Pongrácz
3,877625
answered Aug 17 at 7:00
A. Pongrácz
3,877625
3,877625
add a comment |Â
add a comment |Â
up vote
2
down vote
By L'Hospital we have
$lim_x to 1frace^2xlogxx-1= lim_x to 1(2e^2x log x+e^2xfrac1x)=e^2$.
answered Aug 17 at 7:13
Fred
38.2k1238
add a comment |Â
up vote
2
down vote
By L'Hospital we have
$lim_x to 1frace^2xlogxx-1= lim_x to 1(2e^2x log x+e^2xfrac1x)=e^2$.
answered Aug 17 at 7:13
Fred
38.2k1238
add a comment |Â
up vote
2
down vote
up vote
2
down vote
By L'Hospital we have
$lim_x to 1frace^2xlogxx-1= lim_x to 1(2e^2x log x+e^2xfrac1x)=e^2$.
answered Aug 17 at 7:13
Fred
38.2k1238
By L'Hospital we have
$lim_x to 1frace^2xlogxx-1= lim_x to 1(2e^2x log x+e^2xfrac1x)=e^2$.
answered Aug 17 at 7:13
Fred
38.2k1238
answered Aug 17 at 7:13
Fred
38.2k1238
answered Aug 17 at 7:13
Fred
38.2k1238
answered Aug 17 at 7:13
Fred
38.2k1238
38.2k1238
add a comment |Â
add a comment |Â
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That is a mistake. If by textbook you mean lecture notes, you should point this out to the professor who wrote it. It will be appreciated. (If it is a proper book, you can also write an e-mail to the author.)
– A. Pongrácz
Aug 17 at 8:23