Why isn't L Hospital Rule applicable here?
$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$
I'm trying to use this form of LH Rule. 
calculus
asked 5 hours ago
user1752323
324
add a comment |
$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$
I'm trying to use this form of LH Rule.
calculus
asked 5 hours ago
user1752323
324
Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago
6
The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago
You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago
@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
40 mins ago
add a comment |
$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$
I'm trying to use this form of LH Rule.
calculus
asked 5 hours ago
user1752323
324
$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$
I'm trying to use this form of LH Rule.
calculus
calculus
asked 5 hours ago
user1752323
324
asked 5 hours ago
user1752323
324
asked 5 hours ago
user1752323
324
asked 5 hours ago
user1752323
324
asked 5 hours ago
user1752323
324
324
Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago
6
The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago
You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago
@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
40 mins ago
add a comment |
Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago
6
The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago
You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago
@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
40 mins ago
Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago
Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago
6
6
The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago
The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago
You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago
You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago
@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
40 mins ago
@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
40 mins ago
add a comment |
3 Answers
3
active
oldest
votes
You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.
The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.
More generally you need $infty/infty$ or $0/0$ on either side.
If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$
answered 4 hours ago
egreg
178k1484201
add a comment |
L'Hopital's rule does not apply because the limit
$$
lim_{xto 0}e^{1/x}
$$
does not exist, since
$$
lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
$$
So, consequently
$$
lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
$$
edited 4 hours ago
Szeto
6,4362926
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
2
This is not the reason why l'Hôpital cannot be applied.
– egreg
4 hours ago
add a comment |
here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
$$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
So,
$$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$
so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.
answered 4 hours ago
Rakibul Islam Prince
952211
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.
The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.
More generally you need $infty/infty$ or $0/0$ on either side.
If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$
answered 4 hours ago
egreg
178k1484201
add a comment |
You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.
The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.
More generally you need $infty/infty$ or $0/0$ on either side.
If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$
answered 4 hours ago
egreg
178k1484201
add a comment |
You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.
The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.
More generally you need $infty/infty$ or $0/0$ on either side.
If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$
answered 4 hours ago
egreg
178k1484201
You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.
The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.
More generally you need $infty/infty$ or $0/0$ on either side.
If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$
answered 4 hours ago
egreg
178k1484201
edited 4 hours ago
answered 4 hours ago
egreg
178k1484201
answered 4 hours ago
egreg
178k1484201
answered 4 hours ago
egreg
178k1484201
178k1484201
add a comment |
add a comment |
L'Hopital's rule does not apply because the limit
$$
lim_{xto 0}e^{1/x}
$$
does not exist, since
$$
lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
$$
So, consequently
$$
lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
$$
edited 4 hours ago
Szeto
6,4362926
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
2
This is not the reason why l'Hôpital cannot be applied.
– egreg
4 hours ago
add a comment |
L'Hopital's rule does not apply because the limit
$$
lim_{xto 0}e^{1/x}
$$
does not exist, since
$$
lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
$$
So, consequently
$$
lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
$$
edited 4 hours ago
Szeto
6,4362926
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
2
This is not the reason why l'Hôpital cannot be applied.
– egreg
4 hours ago
add a comment |
L'Hopital's rule does not apply because the limit
$$
lim_{xto 0}e^{1/x}
$$
does not exist, since
$$
lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
$$
So, consequently
$$
lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
$$
edited 4 hours ago
Szeto
6,4362926
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
L'Hopital's rule does not apply because the limit
$$
lim_{xto 0}e^{1/x}
$$
does not exist, since
$$
lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
$$
So, consequently
$$
lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
$$
edited 4 hours ago
Szeto
6,4362926
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
edited 4 hours ago
Szeto
6,4362926
edited 4 hours ago
Szeto
6,4362926
edited 4 hours ago
Szeto
6,4362926
6,4362926
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
answered 5 hours ago
Yiorgos S. Smyrlis
62.6k1383163
62.6k1383163
2
This is not the reason why l'Hôpital cannot be applied.
– egreg
4 hours ago
add a comment |
2
This is not the reason why l'Hôpital cannot be applied.
– egreg
4 hours ago
2
2
This is not the reason why l'Hôpital cannot be applied.
– egreg
4 hours ago
This is not the reason why l'Hôpital cannot be applied.
– egreg
4 hours ago
add a comment |
here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
$$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
So,
$$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$
so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.
answered 4 hours ago
Rakibul Islam Prince
952211
add a comment |
here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
$$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
So,
$$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$
so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.
answered 4 hours ago
Rakibul Islam Prince
952211
add a comment |
here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
$$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
So,
$$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$
so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.
answered 4 hours ago
Rakibul Islam Prince
952211
here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
$$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
So,
$$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$
so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.
answered 4 hours ago
Rakibul Islam Prince
952211
answered 4 hours ago
Rakibul Islam Prince
952211
answered 4 hours ago
Rakibul Islam Prince
952211
answered 4 hours ago
Rakibul Islam Prince
952211
952211
add a comment |
add a comment |
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Why not? I think it is allright
– José Alejandro Aburto Araneda
5 hours ago
6
The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
5 hours ago
You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
4 hours ago
@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
40 mins ago