limit as x approaches negative infinity of ln x [why the answer is infinity when x>0] [on hold]
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limit as $x$ approaches negative infinity of $ln x$ [why the answer is infinity when $x > 0$]
$$lim _{xto -infty }bigl(ln (x)bigr)=infty $$
Somebody help me.
limits
edited Nov 20 at 0:21
Bernard
115k637107
asked Nov 20 at 0:15
Liquidice Slayer
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put on hold as off-topic by amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R Nov 20 at 2:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-3
down vote
favorite
limit as $x$ approaches negative infinity of $ln x$ [why the answer is infinity when $x > 0$]
$$lim _{xto -infty }bigl(ln (x)bigr)=infty $$
Somebody help me.
limits
edited Nov 20 at 0:21
Bernard
115k637107
asked Nov 20 at 0:15
Liquidice Slayer
1
New contributor
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R Nov 20 at 2:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
That is simply because in the real domain, the logarithm is defined only for $x>0$.
– Bernard
Nov 20 at 0:22
so won't the limit be dne
– Liquidice Slayer
Nov 20 at 0:23
"DNE" is a context-dependent definition. It is not incorrect to say the limit does not exist if the limit is infinity, because typically we only say a limit converges to a "real number" (i.e. something like $1, 5.5, pi...$, etc.). $infty$ is not a real number, so in that way we can say the limit does not exist - because it does not exist in the reals. We may specify $+infty$ or $-infty$ to clarify the behavior of the function - or some functions simply do not converge to a limit whatsoever, finite or infinite, and we can say the limit does not exist in those cases too.
– Eevee Trainer
Nov 20 at 0:25
Negative numbers are not defined in ln(x) so the limit to a negative number won't exist. At zero, its negative infinity, why does going further magically flip it to infinity? The limit as x approaches -1 is undefined. So why would - infinity be infinity.
– Liquidice Slayer
Nov 20 at 0:29
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
limit as $x$ approaches negative infinity of $ln x$ [why the answer is infinity when $x > 0$]
$$lim _{xto -infty }bigl(ln (x)bigr)=infty $$
Somebody help me.
limits
edited Nov 20 at 0:21
Bernard
115k637107
asked Nov 20 at 0:15
Liquidice Slayer
1
New contributor
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
limit as $x$ approaches negative infinity of $ln x$ [why the answer is infinity when $x > 0$]
$$lim _{xto -infty }bigl(ln (x)bigr)=infty $$
Somebody help me.
limits
limits
edited Nov 20 at 0:21
Bernard
115k637107
asked Nov 20 at 0:15
Liquidice Slayer
1
New contributor
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 20 at 0:21
Bernard
115k637107
asked Nov 20 at 0:15
Liquidice Slayer
1
New contributor
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 20 at 0:21
Bernard
115k637107
edited Nov 20 at 0:21
Bernard
115k637107
edited Nov 20 at 0:21
Bernard
115k637107
115k637107
asked Nov 20 at 0:15
Liquidice Slayer
1
New contributor
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 0:15
Liquidice Slayer
1
asked Nov 20 at 0:15
Liquidice Slayer
1
1
New contributor
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Liquidice Slayer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R Nov 20 at 2:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R Nov 20 at 2:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Kavi Rama Murthy, Cesareo, Shailesh, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
That is simply because in the real domain, the logarithm is defined only for $x>0$.
– Bernard
Nov 20 at 0:22
so won't the limit be dne
– Liquidice Slayer
Nov 20 at 0:23
"DNE" is a context-dependent definition. It is not incorrect to say the limit does not exist if the limit is infinity, because typically we only say a limit converges to a "real number" (i.e. something like $1, 5.5, pi...$, etc.). $infty$ is not a real number, so in that way we can say the limit does not exist - because it does not exist in the reals. We may specify $+infty$ or $-infty$ to clarify the behavior of the function - or some functions simply do not converge to a limit whatsoever, finite or infinite, and we can say the limit does not exist in those cases too.
– Eevee Trainer
Nov 20 at 0:25
Negative numbers are not defined in ln(x) so the limit to a negative number won't exist. At zero, its negative infinity, why does going further magically flip it to infinity? The limit as x approaches -1 is undefined. So why would - infinity be infinity.
– Liquidice Slayer
Nov 20 at 0:29
add a comment |
That is simply because in the real domain, the logarithm is defined only for $x>0$.
– Bernard
Nov 20 at 0:22
so won't the limit be dne
– Liquidice Slayer
Nov 20 at 0:23
"DNE" is a context-dependent definition. It is not incorrect to say the limit does not exist if the limit is infinity, because typically we only say a limit converges to a "real number" (i.e. something like $1, 5.5, pi...$, etc.). $infty$ is not a real number, so in that way we can say the limit does not exist - because it does not exist in the reals. We may specify $+infty$ or $-infty$ to clarify the behavior of the function - or some functions simply do not converge to a limit whatsoever, finite or infinite, and we can say the limit does not exist in those cases too.
– Eevee Trainer
Nov 20 at 0:25
Negative numbers are not defined in ln(x) so the limit to a negative number won't exist. At zero, its negative infinity, why does going further magically flip it to infinity? The limit as x approaches -1 is undefined. So why would - infinity be infinity.
– Liquidice Slayer
Nov 20 at 0:29
That is simply because in the real domain, the logarithm is defined only for $x>0$.
– Bernard
Nov 20 at 0:22
That is simply because in the real domain, the logarithm is defined only for $x>0$.
– Bernard
Nov 20 at 0:22
so won't the limit be dne
– Liquidice Slayer
Nov 20 at 0:23
so won't the limit be dne
– Liquidice Slayer
Nov 20 at 0:23
"DNE" is a context-dependent definition. It is not incorrect to say the limit does not exist if the limit is infinity, because typically we only say a limit converges to a "real number" (i.e. something like $1, 5.5, pi...$, etc.). $infty$ is not a real number, so in that way we can say the limit does not exist - because it does not exist in the reals. We may specify $+infty$ or $-infty$ to clarify the behavior of the function - or some functions simply do not converge to a limit whatsoever, finite or infinite, and we can say the limit does not exist in those cases too.
– Eevee Trainer
Nov 20 at 0:25
"DNE" is a context-dependent definition. It is not incorrect to say the limit does not exist if the limit is infinity, because typically we only say a limit converges to a "real number" (i.e. something like $1, 5.5, pi...$, etc.). $infty$ is not a real number, so in that way we can say the limit does not exist - because it does not exist in the reals. We may specify $+infty$ or $-infty$ to clarify the behavior of the function - or some functions simply do not converge to a limit whatsoever, finite or infinite, and we can say the limit does not exist in those cases too.
– Eevee Trainer
Nov 20 at 0:25
Negative numbers are not defined in ln(x) so the limit to a negative number won't exist. At zero, its negative infinity, why does going further magically flip it to infinity? The limit as x approaches -1 is undefined. So why would - infinity be infinity.
– Liquidice Slayer
Nov 20 at 0:29
Negative numbers are not defined in ln(x) so the limit to a negative number won't exist. At zero, its negative infinity, why does going further magically flip it to infinity? The limit as x approaches -1 is undefined. So why would - infinity be infinity.
– Liquidice Slayer
Nov 20 at 0:29
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Could you clarify the question a little more as what I am seeing now is $displaystyle lim_{xto -infty} ln(x) $? Again, the function $ln(x)$ is defined for $x ge 0$ so there is no limit when x is negative. For the case when $x to +infty$, you need to show that $f(x) = ln(x)$ is the increasing function and it will explain why the limit goes to infinity.
answered Nov 20 at 0:33
Viet Hoang Quoc
544
No dude, there is a limit. The limit is infinity. I just need to know why.
– Liquidice Slayer
Nov 20 at 0:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Could you clarify the question a little more as what I am seeing now is $displaystyle lim_{xto -infty} ln(x) $? Again, the function $ln(x)$ is defined for $x ge 0$ so there is no limit when x is negative. For the case when $x to +infty$, you need to show that $f(x) = ln(x)$ is the increasing function and it will explain why the limit goes to infinity.
answered Nov 20 at 0:33
Viet Hoang Quoc
544
No dude, there is a limit. The limit is infinity. I just need to know why.
– Liquidice Slayer
Nov 20 at 0:35
add a comment |
up vote
0
down vote
Could you clarify the question a little more as what I am seeing now is $displaystyle lim_{xto -infty} ln(x) $? Again, the function $ln(x)$ is defined for $x ge 0$ so there is no limit when x is negative. For the case when $x to +infty$, you need to show that $f(x) = ln(x)$ is the increasing function and it will explain why the limit goes to infinity.
answered Nov 20 at 0:33
Viet Hoang Quoc
544
No dude, there is a limit. The limit is infinity. I just need to know why.
– Liquidice Slayer
Nov 20 at 0:35
add a comment |
up vote
0
down vote
up vote
0
down vote
Could you clarify the question a little more as what I am seeing now is $displaystyle lim_{xto -infty} ln(x) $? Again, the function $ln(x)$ is defined for $x ge 0$ so there is no limit when x is negative. For the case when $x to +infty$, you need to show that $f(x) = ln(x)$ is the increasing function and it will explain why the limit goes to infinity.
answered Nov 20 at 0:33
Viet Hoang Quoc
544
Could you clarify the question a little more as what I am seeing now is $displaystyle lim_{xto -infty} ln(x) $? Again, the function $ln(x)$ is defined for $x ge 0$ so there is no limit when x is negative. For the case when $x to +infty$, you need to show that $f(x) = ln(x)$ is the increasing function and it will explain why the limit goes to infinity.
answered Nov 20 at 0:33
Viet Hoang Quoc
544
answered Nov 20 at 0:33
Viet Hoang Quoc
544
answered Nov 20 at 0:33
Viet Hoang Quoc
544
answered Nov 20 at 0:33
Viet Hoang Quoc
544
544
No dude, there is a limit. The limit is infinity. I just need to know why.
– Liquidice Slayer
Nov 20 at 0:35
add a comment |
No dude, there is a limit. The limit is infinity. I just need to know why.
– Liquidice Slayer
Nov 20 at 0:35
No dude, there is a limit. The limit is infinity. I just need to know why.
– Liquidice Slayer
Nov 20 at 0:35
No dude, there is a limit. The limit is infinity. I just need to know why.
– Liquidice Slayer
Nov 20 at 0:35
add a comment |
That is simply because in the real domain, the logarithm is defined only for $x>0$.
– Bernard
Nov 20 at 0:22
so won't the limit be dne
– Liquidice Slayer
Nov 20 at 0:23
"DNE" is a context-dependent definition. It is not incorrect to say the limit does not exist if the limit is infinity, because typically we only say a limit converges to a "real number" (i.e. something like $1, 5.5, pi...$, etc.). $infty$ is not a real number, so in that way we can say the limit does not exist - because it does not exist in the reals. We may specify $+infty$ or $-infty$ to clarify the behavior of the function - or some functions simply do not converge to a limit whatsoever, finite or infinite, and we can say the limit does not exist in those cases too.
– Eevee Trainer
Nov 20 at 0:25
Negative numbers are not defined in ln(x) so the limit to a negative number won't exist. At zero, its negative infinity, why does going further magically flip it to infinity? The limit as x approaches -1 is undefined. So why would - infinity be infinity.
– Liquidice Slayer
Nov 20 at 0:29