Mixing
How to deal with $g(f(x))$ in the limit?
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Let $limlimits_x to infty f(x)=infty$. Does the following hold as $xtoinfty$?:
$$exp(x) leq expleft(fracx1+1/ln(1+f(x))right)$$
My effort: If above inequality holds in the limit, the it must be that the following holds in the limit (take $log$ of both sides):
$$x leq fracx1+1/ln(1+f(x))$$
Or, equivalenty:
$$1/ln(1+f(x)) leq 0$$
which is valid since $f(x)to infty$ (given in the problem statement), and thus $1/ln(1+f(x)) to 0$. So, the inequality in the problem statement holds in the limit.
Is this thought process correct? If not, please provide me with a correct proof if the inequality above holds or does not hold for all $f(x)$ that tend to infinity as $x to infty$.
limits limits-without-lhopital
asked 2 days ago
Susan_Math123
825319
add a comment |Â
up vote
1
down vote
favorite
Let $limlimits_x to infty f(x)=infty$. Does the following hold as $xtoinfty$?:
$$exp(x) leq expleft(fracx1+1/ln(1+f(x))right)$$
My effort: If above inequality holds in the limit, the it must be that the following holds in the limit (take $log$ of both sides):
$$x leq fracx1+1/ln(1+f(x))$$
Or, equivalenty:
$$1/ln(1+f(x)) leq 0$$
which is valid since $f(x)to infty$ (given in the problem statement), and thus $1/ln(1+f(x)) to 0$. So, the inequality in the problem statement holds in the limit.
Is this thought process correct? If not, please provide me with a correct proof if the inequality above holds or does not hold for all $f(x)$ that tend to infinity as $x to infty$.
limits limits-without-lhopital
asked 2 days ago
Susan_Math123
825319
1
I think I understand what you're confused about. Usually, when people talk about an inequality holding in the limit, what the mean is that it is true for all sufficiently large $x$, which you can easily show is false for this problem
– Dark Malthorp
2 days ago
@DarkMalthorp Thanks for the clarification.
– Susan_Math123
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $limlimits_x to infty f(x)=infty$. Does the following hold as $xtoinfty$?:
$$exp(x) leq expleft(fracx1+1/ln(1+f(x))right)$$
My effort: If above inequality holds in the limit, the it must be that the following holds in the limit (take $log$ of both sides):
$$x leq fracx1+1/ln(1+f(x))$$
Or, equivalenty:
$$1/ln(1+f(x)) leq 0$$
which is valid since $f(x)to infty$ (given in the problem statement), and thus $1/ln(1+f(x)) to 0$. So, the inequality in the problem statement holds in the limit.
Is this thought process correct? If not, please provide me with a correct proof if the inequality above holds or does not hold for all $f(x)$ that tend to infinity as $x to infty$.
limits limits-without-lhopital
asked 2 days ago
Susan_Math123
825319
Let $limlimits_x to infty f(x)=infty$. Does the following hold as $xtoinfty$?:
$$exp(x) leq expleft(fracx1+1/ln(1+f(x))right)$$
My effort: If above inequality holds in the limit, the it must be that the following holds in the limit (take $log$ of both sides):
$$x leq fracx1+1/ln(1+f(x))$$
Or, equivalenty:
$$1/ln(1+f(x)) leq 0$$
which is valid since $f(x)to infty$ (given in the problem statement), and thus $1/ln(1+f(x)) to 0$. So, the inequality in the problem statement holds in the limit.
Is this thought process correct? If not, please provide me with a correct proof if the inequality above holds or does not hold for all $f(x)$ that tend to infinity as $x to infty$.
limits limits-without-lhopital
asked 2 days ago
Susan_Math123
825319
edited 2 days ago
asked 2 days ago
Susan_Math123
825319
asked 2 days ago
Susan_Math123
825319
asked 2 days ago
Susan_Math123
825319
825319
1
I think I understand what you're confused about. Usually, when people talk about an inequality holding in the limit, what the mean is that it is true for all sufficiently large $x$, which you can easily show is false for this problem
– Dark Malthorp
2 days ago
@DarkMalthorp Thanks for the clarification.
– Susan_Math123
2 days ago
add a comment |Â
1
I think I understand what you're confused about. Usually, when people talk about an inequality holding in the limit, what the mean is that it is true for all sufficiently large $x$, which you can easily show is false for this problem
– Dark Malthorp
2 days ago
@DarkMalthorp Thanks for the clarification.
– Susan_Math123
2 days ago
1
1
I think I understand what you're confused about. Usually, when people talk about an inequality holding in the limit, what the mean is that it is true for all sufficiently large $x$, which you can easily show is false for this problem
– Dark Malthorp
2 days ago
I think I understand what you're confused about. Usually, when people talk about an inequality holding in the limit, what the mean is that it is true for all sufficiently large $x$, which you can easily show is false for this problem
– Dark Malthorp
2 days ago
@DarkMalthorp Thanks for the clarification.
– Susan_Math123
2 days ago
@DarkMalthorp Thanks for the clarification.
– Susan_Math123
2 days ago
add a comment |Â
4 Answers
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up vote
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accepted
We have that, since exp function is increasing
$$e^x leq e^left(fracx1+1/ln(1+f(x))right)iff x leq fracx1+1/ln(1+f(x))$$
and since eventually $x>0$, $1+f(x)>1$ we have
$$x leq fracx1+1/ln(1+f(x)) iff frac1ln(1+f(x)) le 0$$
which is not true.
answered yesterday
gimusi
63.6k73380
add a comment |Â
up vote
1
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So the problem with your reasoning comes when you claim
$$
1/log(1+f(x)) le 0
$$
Since $f(x) rightarrow infty$, this means that $f(x) > 0$ for all $x$ sufficiently large. When $f(x) > 0$ however, we would have that $1/log(1+f(x)) >0$ because $log(z) > 0$ whenever $z > 1$.
All the other steps in your reasoning are correct, so in fact the answer is that the inequality is false for large $x$.
The point you seem to be confused on is the fact that
$$
lim_xrightarrowinfty 1/log(1 + f(x)) le 0
$$
which is true, but is not good enough to make the original inequality true. Why? Because both sides go to infinity, so even though the $1/log(1 + f(x))$ is trying to go to $0$, since it is strictly positive it can still affect the rate at which the function $exp(fracx1 + 1/log(1 + f(x)))$. For example, suppose $f(x) = x - 1$. Clearly, that goes to infinity. You would therefore be claiming that in the limit
$$
exp(x) le expleft(fracx1 + 1/log(1+f(x))right) = expleft(fracx1 + 1/log(x)right) = expleft(fracxlog(x)log(x) + 1right)
$$
but the left hand side goes to infinity much faster than the left. If we take the ratio of both sides, you would claim:
begineqnarray
fracexp(x)expleft(fracxlog(x)log(x) + 1right) &le& 1 \
expleft(x - fracxlog(x)log(x) + 1right) &le& 1 \
expleft(xleft(1 - fraclog(x)log(x) + 1right)right) &le& 1 \
expleft(xleft(frac1log(x) + 1right)right) &le& 1
endeqnarray
but the right hand side goes to infinity in the limit.
answered 2 days ago
Dark Malthorp
667111
Thanks. If you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Just use your same argument until you get to the step $1/log(1 + f(x)) le 0$.
– Dark Malthorp
2 days ago
what should I do instead of the step you mentioned?
– Susan_Math123
2 days ago
1
The limit of $f(x)$ being positive infinity implies that $1/log(1 + f(x)) > 0$ for large $x$ so the original inequality is false because it would imply $1/log(1+f(x)) le 0$
– Dark Malthorp
2 days ago
I agree, $1/log(1+f(x))>0$, but $1/log(1+f(x))$ can get as close as desired to zero, and it becomes zero as $x to infty$. That fixes it.
– Susan_Math123
2 days ago
 |Â
show 3 more comments
up vote
0
down vote
Your thought process is rather tricky, as your last line holds exactly in the limit of $xtoinfty$, and is wrong for all other $x>0$.
You however start your deduction from the desired result and move backwards - which is okay, as every step you make is bijective for sufficiently big $x$, but to make conclusions about the bijectivity of your steps regarding a limit, you need to prove that the limit exists (remember, all those rules one learns require this).
This is not given in your proof, as e.g. $lim_xtoinfty exp(x) = infty$.
answered 2 days ago
Sudix
7891216
Thanks. But, can you tell me more clearly? I only need it to hold when $x$ tends to infinity.
– Susan_Math123
2 days ago
And your argumentation doesn't provide this result.
– Sudix
2 days ago
add a comment |Â
up vote
0
down vote
Since all of this process is reversible so your method is right but we have $$dfrac1ln1+f(x)le 0leftarrowrightarrowln1+f(x)< 0$$which means that $$f(x)<0$$so it doesn't hold under your constraints.
answered 2 days ago
Mostafa Ayaz
8,5203530
Thanks, but the right side of your first equation is missing "f(x) to infty". Wouldn't that fix my argument?
– Susan_Math123
2 days ago
Also, if you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Your way is correct and this question would be satisfied using your constraint if $$lim_xto -inftyf(x)=infty$$
– Mostafa Ayaz
2 days ago
I have said what you need at the beginning of my problem statement.
– Susan_Math123
2 days ago
1
Yes absolutely but take for example $f(x)=e^x-1$ then your inequality reduces to $$e^xle e^fracx^2x+1$$which is absolutely wrong for large $x$
– Mostafa Ayaz
2 days ago
 |Â
show 1 more comment
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have that, since exp function is increasing
$$e^x leq e^left(fracx1+1/ln(1+f(x))right)iff x leq fracx1+1/ln(1+f(x))$$
and since eventually $x>0$, $1+f(x)>1$ we have
$$x leq fracx1+1/ln(1+f(x)) iff frac1ln(1+f(x)) le 0$$
which is not true.
answered yesterday
gimusi
63.6k73380
add a comment |Â
up vote
1
down vote
accepted
We have that, since exp function is increasing
$$e^x leq e^left(fracx1+1/ln(1+f(x))right)iff x leq fracx1+1/ln(1+f(x))$$
and since eventually $x>0$, $1+f(x)>1$ we have
$$x leq fracx1+1/ln(1+f(x)) iff frac1ln(1+f(x)) le 0$$
which is not true.
answered yesterday
gimusi
63.6k73380
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have that, since exp function is increasing
$$e^x leq e^left(fracx1+1/ln(1+f(x))right)iff x leq fracx1+1/ln(1+f(x))$$
and since eventually $x>0$, $1+f(x)>1$ we have
$$x leq fracx1+1/ln(1+f(x)) iff frac1ln(1+f(x)) le 0$$
which is not true.
answered yesterday
gimusi
63.6k73380
We have that, since exp function is increasing
$$e^x leq e^left(fracx1+1/ln(1+f(x))right)iff x leq fracx1+1/ln(1+f(x))$$
and since eventually $x>0$, $1+f(x)>1$ we have
$$x leq fracx1+1/ln(1+f(x)) iff frac1ln(1+f(x)) le 0$$
which is not true.
answered yesterday
gimusi
63.6k73380
answered yesterday
gimusi
63.6k73380
answered yesterday
gimusi
63.6k73380
answered yesterday
gimusi
63.6k73380
63.6k73380
add a comment |Â
add a comment |Â
up vote
1
down vote
So the problem with your reasoning comes when you claim
$$
1/log(1+f(x)) le 0
$$
Since $f(x) rightarrow infty$, this means that $f(x) > 0$ for all $x$ sufficiently large. When $f(x) > 0$ however, we would have that $1/log(1+f(x)) >0$ because $log(z) > 0$ whenever $z > 1$.
All the other steps in your reasoning are correct, so in fact the answer is that the inequality is false for large $x$.
The point you seem to be confused on is the fact that
$$
lim_xrightarrowinfty 1/log(1 + f(x)) le 0
$$
which is true, but is not good enough to make the original inequality true. Why? Because both sides go to infinity, so even though the $1/log(1 + f(x))$ is trying to go to $0$, since it is strictly positive it can still affect the rate at which the function $exp(fracx1 + 1/log(1 + f(x)))$. For example, suppose $f(x) = x - 1$. Clearly, that goes to infinity. You would therefore be claiming that in the limit
$$
exp(x) le expleft(fracx1 + 1/log(1+f(x))right) = expleft(fracx1 + 1/log(x)right) = expleft(fracxlog(x)log(x) + 1right)
$$
but the left hand side goes to infinity much faster than the left. If we take the ratio of both sides, you would claim:
begineqnarray
fracexp(x)expleft(fracxlog(x)log(x) + 1right) &le& 1 \
expleft(x - fracxlog(x)log(x) + 1right) &le& 1 \
expleft(xleft(1 - fraclog(x)log(x) + 1right)right) &le& 1 \
expleft(xleft(frac1log(x) + 1right)right) &le& 1
endeqnarray
but the right hand side goes to infinity in the limit.
answered 2 days ago
Dark Malthorp
667111
Thanks. If you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Just use your same argument until you get to the step $1/log(1 + f(x)) le 0$.
– Dark Malthorp
2 days ago
what should I do instead of the step you mentioned?
– Susan_Math123
2 days ago
1
The limit of $f(x)$ being positive infinity implies that $1/log(1 + f(x)) > 0$ for large $x$ so the original inequality is false because it would imply $1/log(1+f(x)) le 0$
– Dark Malthorp
2 days ago
I agree, $1/log(1+f(x))>0$, but $1/log(1+f(x))$ can get as close as desired to zero, and it becomes zero as $x to infty$. That fixes it.
– Susan_Math123
2 days ago
 |Â
show 3 more comments
up vote
1
down vote
So the problem with your reasoning comes when you claim
$$
1/log(1+f(x)) le 0
$$
Since $f(x) rightarrow infty$, this means that $f(x) > 0$ for all $x$ sufficiently large. When $f(x) > 0$ however, we would have that $1/log(1+f(x)) >0$ because $log(z) > 0$ whenever $z > 1$.
All the other steps in your reasoning are correct, so in fact the answer is that the inequality is false for large $x$.
The point you seem to be confused on is the fact that
$$
lim_xrightarrowinfty 1/log(1 + f(x)) le 0
$$
which is true, but is not good enough to make the original inequality true. Why? Because both sides go to infinity, so even though the $1/log(1 + f(x))$ is trying to go to $0$, since it is strictly positive it can still affect the rate at which the function $exp(fracx1 + 1/log(1 + f(x)))$. For example, suppose $f(x) = x - 1$. Clearly, that goes to infinity. You would therefore be claiming that in the limit
$$
exp(x) le expleft(fracx1 + 1/log(1+f(x))right) = expleft(fracx1 + 1/log(x)right) = expleft(fracxlog(x)log(x) + 1right)
$$
but the left hand side goes to infinity much faster than the left. If we take the ratio of both sides, you would claim:
begineqnarray
fracexp(x)expleft(fracxlog(x)log(x) + 1right) &le& 1 \
expleft(x - fracxlog(x)log(x) + 1right) &le& 1 \
expleft(xleft(1 - fraclog(x)log(x) + 1right)right) &le& 1 \
expleft(xleft(frac1log(x) + 1right)right) &le& 1
endeqnarray
but the right hand side goes to infinity in the limit.
answered 2 days ago
Dark Malthorp
667111
Thanks. If you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Just use your same argument until you get to the step $1/log(1 + f(x)) le 0$.
– Dark Malthorp
2 days ago
what should I do instead of the step you mentioned?
– Susan_Math123
2 days ago
1
The limit of $f(x)$ being positive infinity implies that $1/log(1 + f(x)) > 0$ for large $x$ so the original inequality is false because it would imply $1/log(1+f(x)) le 0$
– Dark Malthorp
2 days ago
I agree, $1/log(1+f(x))>0$, but $1/log(1+f(x))$ can get as close as desired to zero, and it becomes zero as $x to infty$. That fixes it.
– Susan_Math123
2 days ago
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
So the problem with your reasoning comes when you claim
$$
1/log(1+f(x)) le 0
$$
Since $f(x) rightarrow infty$, this means that $f(x) > 0$ for all $x$ sufficiently large. When $f(x) > 0$ however, we would have that $1/log(1+f(x)) >0$ because $log(z) > 0$ whenever $z > 1$.
All the other steps in your reasoning are correct, so in fact the answer is that the inequality is false for large $x$.
The point you seem to be confused on is the fact that
$$
lim_xrightarrowinfty 1/log(1 + f(x)) le 0
$$
which is true, but is not good enough to make the original inequality true. Why? Because both sides go to infinity, so even though the $1/log(1 + f(x))$ is trying to go to $0$, since it is strictly positive it can still affect the rate at which the function $exp(fracx1 + 1/log(1 + f(x)))$. For example, suppose $f(x) = x - 1$. Clearly, that goes to infinity. You would therefore be claiming that in the limit
$$
exp(x) le expleft(fracx1 + 1/log(1+f(x))right) = expleft(fracx1 + 1/log(x)right) = expleft(fracxlog(x)log(x) + 1right)
$$
but the left hand side goes to infinity much faster than the left. If we take the ratio of both sides, you would claim:
begineqnarray
fracexp(x)expleft(fracxlog(x)log(x) + 1right) &le& 1 \
expleft(x - fracxlog(x)log(x) + 1right) &le& 1 \
expleft(xleft(1 - fraclog(x)log(x) + 1right)right) &le& 1 \
expleft(xleft(frac1log(x) + 1right)right) &le& 1
endeqnarray
but the right hand side goes to infinity in the limit.
answered 2 days ago
Dark Malthorp
667111
So the problem with your reasoning comes when you claim
$$
1/log(1+f(x)) le 0
$$
Since $f(x) rightarrow infty$, this means that $f(x) > 0$ for all $x$ sufficiently large. When $f(x) > 0$ however, we would have that $1/log(1+f(x)) >0$ because $log(z) > 0$ whenever $z > 1$.
All the other steps in your reasoning are correct, so in fact the answer is that the inequality is false for large $x$.
The point you seem to be confused on is the fact that
$$
lim_xrightarrowinfty 1/log(1 + f(x)) le 0
$$
which is true, but is not good enough to make the original inequality true. Why? Because both sides go to infinity, so even though the $1/log(1 + f(x))$ is trying to go to $0$, since it is strictly positive it can still affect the rate at which the function $exp(fracx1 + 1/log(1 + f(x)))$. For example, suppose $f(x) = x - 1$. Clearly, that goes to infinity. You would therefore be claiming that in the limit
$$
exp(x) le expleft(fracx1 + 1/log(1+f(x))right) = expleft(fracx1 + 1/log(x)right) = expleft(fracxlog(x)log(x) + 1right)
$$
but the left hand side goes to infinity much faster than the left. If we take the ratio of both sides, you would claim:
begineqnarray
fracexp(x)expleft(fracxlog(x)log(x) + 1right) &le& 1 \
expleft(x - fracxlog(x)log(x) + 1right) &le& 1 \
expleft(xleft(1 - fraclog(x)log(x) + 1right)right) &le& 1 \
expleft(xleft(frac1log(x) + 1right)right) &le& 1
endeqnarray
but the right hand side goes to infinity in the limit.
answered 2 days ago
Dark Malthorp
667111
edited 2 days ago
answered 2 days ago
Dark Malthorp
667111
answered 2 days ago
Dark Malthorp
667111
answered 2 days ago
Dark Malthorp
667111
667111
Thanks. If you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Just use your same argument until you get to the step $1/log(1 + f(x)) le 0$.
– Dark Malthorp
2 days ago
what should I do instead of the step you mentioned?
– Susan_Math123
2 days ago
1
The limit of $f(x)$ being positive infinity implies that $1/log(1 + f(x)) > 0$ for large $x$ so the original inequality is false because it would imply $1/log(1+f(x)) le 0$
– Dark Malthorp
2 days ago
I agree, $1/log(1+f(x))>0$, but $1/log(1+f(x))$ can get as close as desired to zero, and it becomes zero as $x to infty$. That fixes it.
– Susan_Math123
2 days ago
 |Â
show 3 more comments
Thanks. If you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Just use your same argument until you get to the step $1/log(1 + f(x)) le 0$.
– Dark Malthorp
2 days ago
what should I do instead of the step you mentioned?
– Susan_Math123
2 days ago
1
The limit of $f(x)$ being positive infinity implies that $1/log(1 + f(x)) > 0$ for large $x$ so the original inequality is false because it would imply $1/log(1+f(x)) le 0$
– Dark Malthorp
2 days ago
I agree, $1/log(1+f(x))>0$, but $1/log(1+f(x))$ can get as close as desired to zero, and it becomes zero as $x to infty$. That fixes it.
– Susan_Math123
2 days ago
Thanks. If you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Thanks. If you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Just use your same argument until you get to the step $1/log(1 + f(x)) le 0$.
– Dark Malthorp
2 days ago
Just use your same argument until you get to the step $1/log(1 + f(x)) le 0$.
– Dark Malthorp
2 days ago
what should I do instead of the step you mentioned?
– Susan_Math123
2 days ago
what should I do instead of the step you mentioned?
– Susan_Math123
2 days ago
1
1
The limit of $f(x)$ being positive infinity implies that $1/log(1 + f(x)) > 0$ for large $x$ so the original inequality is false because it would imply $1/log(1+f(x)) le 0$
– Dark Malthorp
2 days ago
The limit of $f(x)$ being positive infinity implies that $1/log(1 + f(x)) > 0$ for large $x$ so the original inequality is false because it would imply $1/log(1+f(x)) le 0$
– Dark Malthorp
2 days ago
I agree, $1/log(1+f(x))>0$, but $1/log(1+f(x))$ can get as close as desired to zero, and it becomes zero as $x to infty$. That fixes it.
– Susan_Math123
2 days ago
I agree, $1/log(1+f(x))>0$, but $1/log(1+f(x))$ can get as close as desired to zero, and it becomes zero as $x to infty$. That fixes it.
– Susan_Math123
2 days ago
 |Â
show 3 more comments
up vote
0
down vote
Your thought process is rather tricky, as your last line holds exactly in the limit of $xtoinfty$, and is wrong for all other $x>0$.
You however start your deduction from the desired result and move backwards - which is okay, as every step you make is bijective for sufficiently big $x$, but to make conclusions about the bijectivity of your steps regarding a limit, you need to prove that the limit exists (remember, all those rules one learns require this).
This is not given in your proof, as e.g. $lim_xtoinfty exp(x) = infty$.
answered 2 days ago
Sudix
7891216
Thanks. But, can you tell me more clearly? I only need it to hold when $x$ tends to infinity.
– Susan_Math123
2 days ago
And your argumentation doesn't provide this result.
– Sudix
2 days ago
add a comment |Â
up vote
0
down vote
Your thought process is rather tricky, as your last line holds exactly in the limit of $xtoinfty$, and is wrong for all other $x>0$.
You however start your deduction from the desired result and move backwards - which is okay, as every step you make is bijective for sufficiently big $x$, but to make conclusions about the bijectivity of your steps regarding a limit, you need to prove that the limit exists (remember, all those rules one learns require this).
This is not given in your proof, as e.g. $lim_xtoinfty exp(x) = infty$.
answered 2 days ago
Sudix
7891216
Thanks. But, can you tell me more clearly? I only need it to hold when $x$ tends to infinity.
– Susan_Math123
2 days ago
And your argumentation doesn't provide this result.
– Sudix
2 days ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your thought process is rather tricky, as your last line holds exactly in the limit of $xtoinfty$, and is wrong for all other $x>0$.
You however start your deduction from the desired result and move backwards - which is okay, as every step you make is bijective for sufficiently big $x$, but to make conclusions about the bijectivity of your steps regarding a limit, you need to prove that the limit exists (remember, all those rules one learns require this).
This is not given in your proof, as e.g. $lim_xtoinfty exp(x) = infty$.
answered 2 days ago
Sudix
7891216
Your thought process is rather tricky, as your last line holds exactly in the limit of $xtoinfty$, and is wrong for all other $x>0$.
You however start your deduction from the desired result and move backwards - which is okay, as every step you make is bijective for sufficiently big $x$, but to make conclusions about the bijectivity of your steps regarding a limit, you need to prove that the limit exists (remember, all those rules one learns require this).
This is not given in your proof, as e.g. $lim_xtoinfty exp(x) = infty$.
answered 2 days ago
Sudix
7891216
answered 2 days ago
Sudix
7891216
answered 2 days ago
Sudix
7891216
answered 2 days ago
Sudix
7891216
7891216
Thanks. But, can you tell me more clearly? I only need it to hold when $x$ tends to infinity.
– Susan_Math123
2 days ago
And your argumentation doesn't provide this result.
– Sudix
2 days ago
add a comment |Â
Thanks. But, can you tell me more clearly? I only need it to hold when $x$ tends to infinity.
– Susan_Math123
2 days ago
And your argumentation doesn't provide this result.
– Sudix
2 days ago
Thanks. But, can you tell me more clearly? I only need it to hold when $x$ tends to infinity.
– Susan_Math123
2 days ago
Thanks. But, can you tell me more clearly? I only need it to hold when $x$ tends to infinity.
– Susan_Math123
2 days ago
And your argumentation doesn't provide this result.
– Sudix
2 days ago
And your argumentation doesn't provide this result.
– Sudix
2 days ago
add a comment |Â
up vote
0
down vote
Since all of this process is reversible so your method is right but we have $$dfrac1ln1+f(x)le 0leftarrowrightarrowln1+f(x)< 0$$which means that $$f(x)<0$$so it doesn't hold under your constraints.
answered 2 days ago
Mostafa Ayaz
8,5203530
Thanks, but the right side of your first equation is missing "f(x) to infty". Wouldn't that fix my argument?
– Susan_Math123
2 days ago
Also, if you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Your way is correct and this question would be satisfied using your constraint if $$lim_xto -inftyf(x)=infty$$
– Mostafa Ayaz
2 days ago
I have said what you need at the beginning of my problem statement.
– Susan_Math123
2 days ago
1
Yes absolutely but take for example $f(x)=e^x-1$ then your inequality reduces to $$e^xle e^fracx^2x+1$$which is absolutely wrong for large $x$
– Mostafa Ayaz
2 days ago
 |Â
show 1 more comment
up vote
0
down vote
Since all of this process is reversible so your method is right but we have $$dfrac1ln1+f(x)le 0leftarrowrightarrowln1+f(x)< 0$$which means that $$f(x)<0$$so it doesn't hold under your constraints.
answered 2 days ago
Mostafa Ayaz
8,5203530
Thanks, but the right side of your first equation is missing "f(x) to infty". Wouldn't that fix my argument?
– Susan_Math123
2 days ago
Also, if you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Your way is correct and this question would be satisfied using your constraint if $$lim_xto -inftyf(x)=infty$$
– Mostafa Ayaz
2 days ago
I have said what you need at the beginning of my problem statement.
– Susan_Math123
2 days ago
1
Yes absolutely but take for example $f(x)=e^x-1$ then your inequality reduces to $$e^xle e^fracx^2x+1$$which is absolutely wrong for large $x$
– Mostafa Ayaz
2 days ago
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
Since all of this process is reversible so your method is right but we have $$dfrac1ln1+f(x)le 0leftarrowrightarrowln1+f(x)< 0$$which means that $$f(x)<0$$so it doesn't hold under your constraints.
answered 2 days ago
Mostafa Ayaz
8,5203530
Since all of this process is reversible so your method is right but we have $$dfrac1ln1+f(x)le 0leftarrowrightarrowln1+f(x)< 0$$which means that $$f(x)<0$$so it doesn't hold under your constraints.
answered 2 days ago
Mostafa Ayaz
8,5203530
answered 2 days ago
Mostafa Ayaz
8,5203530
answered 2 days ago
Mostafa Ayaz
8,5203530
answered 2 days ago
Mostafa Ayaz
8,5203530
8,5203530
Thanks, but the right side of your first equation is missing "f(x) to infty". Wouldn't that fix my argument?
– Susan_Math123
2 days ago
Also, if you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Your way is correct and this question would be satisfied using your constraint if $$lim_xto -inftyf(x)=infty$$
– Mostafa Ayaz
2 days ago
I have said what you need at the beginning of my problem statement.
– Susan_Math123
2 days ago
1
Yes absolutely but take for example $f(x)=e^x-1$ then your inequality reduces to $$e^xle e^fracx^2x+1$$which is absolutely wrong for large $x$
– Mostafa Ayaz
2 days ago
 |Â
show 1 more comment
Thanks, but the right side of your first equation is missing "f(x) to infty". Wouldn't that fix my argument?
– Susan_Math123
2 days ago
Also, if you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Your way is correct and this question would be satisfied using your constraint if $$lim_xto -inftyf(x)=infty$$
– Mostafa Ayaz
2 days ago
I have said what you need at the beginning of my problem statement.
– Susan_Math123
2 days ago
1
Yes absolutely but take for example $f(x)=e^x-1$ then your inequality reduces to $$e^xle e^fracx^2x+1$$which is absolutely wrong for large $x$
– Mostafa Ayaz
2 days ago
Thanks, but the right side of your first equation is missing "f(x) to infty". Wouldn't that fix my argument?
– Susan_Math123
2 days ago
Thanks, but the right side of your first equation is missing "f(x) to infty". Wouldn't that fix my argument?
– Susan_Math123
2 days ago
Also, if you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Also, if you don' believe my argument is correct, can you provide me with a correct one?
– Susan_Math123
2 days ago
Your way is correct and this question would be satisfied using your constraint if $$lim_xto -inftyf(x)=infty$$
– Mostafa Ayaz
2 days ago
Your way is correct and this question would be satisfied using your constraint if $$lim_xto -inftyf(x)=infty$$
– Mostafa Ayaz
2 days ago
I have said what you need at the beginning of my problem statement.
– Susan_Math123
2 days ago
I have said what you need at the beginning of my problem statement.
– Susan_Math123
2 days ago
1
1
Yes absolutely but take for example $f(x)=e^x-1$ then your inequality reduces to $$e^xle e^fracx^2x+1$$which is absolutely wrong for large $x$
– Mostafa Ayaz
2 days ago
Yes absolutely but take for example $f(x)=e^x-1$ then your inequality reduces to $$e^xle e^fracx^2x+1$$which is absolutely wrong for large $x$
– Mostafa Ayaz
2 days ago
 |Â
show 1 more comment
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1
I think I understand what you're confused about. Usually, when people talk about an inequality holding in the limit, what the mean is that it is true for all sufficiently large $x$, which you can easily show is false for this problem
– Dark Malthorp
2 days ago
@DarkMalthorp Thanks for the clarification.
– Susan_Math123
2 days ago