Mixing
Is it true that $f(x) ge lim_{xto infty} f(x) $?
$begingroup$
Consider a decreasing function $f:[0, infty) tomathbb{R} $. We know that $f(x) le f(0),forall x$, but I was wondering if we can find a lower bound for $f$ using its limit. This is how I came up with the idea that $f(x) ge lim_{xto infty} f(x) $. Intuitively it seems true, but I can't prove it. I tried using the limit's definition, but it didn't help.
EDIT:$lim_{xto infty} f(x)$ exists and if finite.
real-analysis inequality
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
$endgroup$
|
show 1 more comment
$begingroup$
Consider a decreasing function $f:[0, infty) tomathbb{R} $. We know that $f(x) le f(0),forall x$, but I was wondering if we can find a lower bound for $f$ using its limit. This is how I came up with the idea that $f(x) ge lim_{xto infty} f(x) $. Intuitively it seems true, but I can't prove it. I tried using the limit's definition, but it didn't help.
EDIT:$lim_{xto infty} f(x)$ exists and if finite.
real-analysis inequality
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
$endgroup$
$begingroup$
Do you want to assume the limit exists?
$endgroup$
– Randall
Jan 31 at 17:37
$begingroup$
Yes, of course, I forgot to say that.
$endgroup$
– JustAnAmateur
Jan 31 at 17:38
$begingroup$
Thank you guys for your input! I added the constraint that the limit exists.
$endgroup$
– JustAnAmateur
Jan 31 at 17:39
$begingroup$
Since the function is decreasing, then yes, the statement is true.
$endgroup$
– Clayton
Jan 31 at 17:40
1
$begingroup$
Prove it by contradiction.
$endgroup$
– Randall
Jan 31 at 17:40
|
show 1 more comment
$begingroup$
Consider a decreasing function $f:[0, infty) tomathbb{R} $. We know that $f(x) le f(0),forall x$, but I was wondering if we can find a lower bound for $f$ using its limit. This is how I came up with the idea that $f(x) ge lim_{xto infty} f(x) $. Intuitively it seems true, but I can't prove it. I tried using the limit's definition, but it didn't help.
EDIT:$lim_{xto infty} f(x)$ exists and if finite.
real-analysis inequality
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
$endgroup$
Consider a decreasing function $f:[0, infty) tomathbb{R} $. We know that $f(x) le f(0),forall x$, but I was wondering if we can find a lower bound for $f$ using its limit. This is how I came up with the idea that $f(x) ge lim_{xto infty} f(x) $. Intuitively it seems true, but I can't prove it. I tried using the limit's definition, but it didn't help.
EDIT:$lim_{xto infty} f(x)$ exists and if finite.
real-analysis inequality
real-analysis inequality
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
edited Feb 2 at 21:36
DisintegratingByParts
60.3k42681
60.3k42681
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
asked Jan 31 at 17:35
JustAnAmateurJustAnAmateur
1096
1096
$begingroup$
Do you want to assume the limit exists?
$endgroup$
– Randall
Jan 31 at 17:37
$begingroup$
Yes, of course, I forgot to say that.
$endgroup$
– JustAnAmateur
Jan 31 at 17:38
$begingroup$
Thank you guys for your input! I added the constraint that the limit exists.
$endgroup$
– JustAnAmateur
Jan 31 at 17:39
$begingroup$
Since the function is decreasing, then yes, the statement is true.
$endgroup$
– Clayton
Jan 31 at 17:40
1
$begingroup$
Prove it by contradiction.
$endgroup$
– Randall
Jan 31 at 17:40
|
show 1 more comment
$begingroup$
Do you want to assume the limit exists?
$endgroup$
– Randall
Jan 31 at 17:37
$begingroup$
Yes, of course, I forgot to say that.
$endgroup$
– JustAnAmateur
Jan 31 at 17:38
$begingroup$
Thank you guys for your input! I added the constraint that the limit exists.
$endgroup$
– JustAnAmateur
Jan 31 at 17:39
$begingroup$
Since the function is decreasing, then yes, the statement is true.
$endgroup$
– Clayton
Jan 31 at 17:40
1
$begingroup$
Prove it by contradiction.
$endgroup$
– Randall
Jan 31 at 17:40
$begingroup$
Do you want to assume the limit exists?
$endgroup$
– Randall
Jan 31 at 17:37
$begingroup$
Do you want to assume the limit exists?
$endgroup$
– Randall
Jan 31 at 17:37
$begingroup$
Yes, of course, I forgot to say that.
$endgroup$
– JustAnAmateur
Jan 31 at 17:38
$begingroup$
Yes, of course, I forgot to say that.
$endgroup$
– JustAnAmateur
Jan 31 at 17:38
$begingroup$
Thank you guys for your input! I added the constraint that the limit exists.
$endgroup$
– JustAnAmateur
Jan 31 at 17:39
$begingroup$
Thank you guys for your input! I added the constraint that the limit exists.
$endgroup$
– JustAnAmateur
Jan 31 at 17:39
$begingroup$
Since the function is decreasing, then yes, the statement is true.
$endgroup$
– Clayton
Jan 31 at 17:40
$begingroup$
Since the function is decreasing, then yes, the statement is true.
$endgroup$
– Clayton
Jan 31 at 17:40
1
1
$begingroup$
Prove it by contradiction.
$endgroup$
– Randall
Jan 31 at 17:40
$begingroup$
Prove it by contradiction.
$endgroup$
– Randall
Jan 31 at 17:40
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
hint
Let $L$ be the limit.
Assume there exist $age 0$ such that
$$f(a)<L$$
then,
$$forall xge a ;;; f(x)le f(a)<L$$
thus
$$lim_{xto +infty} f(x)le f(a)<L$$
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
(Assuming decreasing means monotonically so, not strictly so. I don't think this matters.)
Let $L = lim_{x to infty} f(x)$. Suppose to the contrary that there is an $x_0$ with $f(x_0) < L$. Since $f$ is decreasing, this means that $f(x) leq f(x_0) < L$ whenever $x geq x_0$. So, "after" $x_0$, $f(x)$ only moves further away from $L$ because it decreases. This will contradict the existence of the limit. This is the intuition.
Pick $epsilon = frac{L-f(x_0)}{3}$. Then there is an $M$ such that if $x geq M$ then $|f(x)-L| < epsilon$. Pick an $x$ greater than both $M$ and $x_0$. Then for this $x$ we have
$$
|f(x)-L| < epsilon
$$
but also, as $f$ decreases,
$$
|f(x) - L| geq |f(x_0)-L| > epsilon
$$
by choice of $epsilon$. Certainly $epsilon < |f(x)-L| < epsilon$ is a contradiction.
answered Jan 31 at 17:57
RandallRandall
10.7k11431
$endgroup$
add a comment |
$begingroup$
Using the limit definition:
Let $displaystyle L = lim_{xtoinfty} f(x)$. Choose $epsilon > 0$. Then there exists $x_0$ such that for all $x ge x_0$ we have $|f(x)-L|<epsilon$. That implies that $f(x) > L - epsilon$ for $x ge x_0$. For $x < x_0$ we have $f(x)ge f(x_0)ge L-epsilon$.
Then $f(x) ge L-epsilon$ for all $x$. As $epsilon >0$ is arbitrary we have $f(x) ge L$ for all $x$.
answered Jan 31 at 18:37
jjagmathjjagmath
3387
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint
Let $L$ be the limit.
Assume there exist $age 0$ such that
$$f(a)<L$$
then,
$$forall xge a ;;; f(x)le f(a)<L$$
thus
$$lim_{xto +infty} f(x)le f(a)<L$$
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
hint
Let $L$ be the limit.
Assume there exist $age 0$ such that
$$f(a)<L$$
then,
$$forall xge a ;;; f(x)le f(a)<L$$
thus
$$lim_{xto +infty} f(x)le f(a)<L$$
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
hint
Let $L$ be the limit.
Assume there exist $age 0$ such that
$$f(a)<L$$
then,
$$forall xge a ;;; f(x)le f(a)<L$$
thus
$$lim_{xto +infty} f(x)le f(a)<L$$
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
hint
Let $L$ be the limit.
Assume there exist $age 0$ such that
$$f(a)<L$$
then,
$$forall xge a ;;; f(x)le f(a)<L$$
thus
$$lim_{xto +infty} f(x)le f(a)<L$$
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Jan 31 at 17:58
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
add a comment |
$begingroup$
(Assuming decreasing means monotonically so, not strictly so. I don't think this matters.)
Let $L = lim_{x to infty} f(x)$. Suppose to the contrary that there is an $x_0$ with $f(x_0) < L$. Since $f$ is decreasing, this means that $f(x) leq f(x_0) < L$ whenever $x geq x_0$. So, "after" $x_0$, $f(x)$ only moves further away from $L$ because it decreases. This will contradict the existence of the limit. This is the intuition.
Pick $epsilon = frac{L-f(x_0)}{3}$. Then there is an $M$ such that if $x geq M$ then $|f(x)-L| < epsilon$. Pick an $x$ greater than both $M$ and $x_0$. Then for this $x$ we have
$$
|f(x)-L| < epsilon
$$
but also, as $f$ decreases,
$$
|f(x) - L| geq |f(x_0)-L| > epsilon
$$
by choice of $epsilon$. Certainly $epsilon < |f(x)-L| < epsilon$ is a contradiction.
answered Jan 31 at 17:57
RandallRandall
10.7k11431
$endgroup$
add a comment |
$begingroup$
(Assuming decreasing means monotonically so, not strictly so. I don't think this matters.)
Let $L = lim_{x to infty} f(x)$. Suppose to the contrary that there is an $x_0$ with $f(x_0) < L$. Since $f$ is decreasing, this means that $f(x) leq f(x_0) < L$ whenever $x geq x_0$. So, "after" $x_0$, $f(x)$ only moves further away from $L$ because it decreases. This will contradict the existence of the limit. This is the intuition.
Pick $epsilon = frac{L-f(x_0)}{3}$. Then there is an $M$ such that if $x geq M$ then $|f(x)-L| < epsilon$. Pick an $x$ greater than both $M$ and $x_0$. Then for this $x$ we have
$$
|f(x)-L| < epsilon
$$
but also, as $f$ decreases,
$$
|f(x) - L| geq |f(x_0)-L| > epsilon
$$
by choice of $epsilon$. Certainly $epsilon < |f(x)-L| < epsilon$ is a contradiction.
answered Jan 31 at 17:57
RandallRandall
10.7k11431
$endgroup$
add a comment |
$begingroup$
(Assuming decreasing means monotonically so, not strictly so. I don't think this matters.)
Let $L = lim_{x to infty} f(x)$. Suppose to the contrary that there is an $x_0$ with $f(x_0) < L$. Since $f$ is decreasing, this means that $f(x) leq f(x_0) < L$ whenever $x geq x_0$. So, "after" $x_0$, $f(x)$ only moves further away from $L$ because it decreases. This will contradict the existence of the limit. This is the intuition.
Pick $epsilon = frac{L-f(x_0)}{3}$. Then there is an $M$ such that if $x geq M$ then $|f(x)-L| < epsilon$. Pick an $x$ greater than both $M$ and $x_0$. Then for this $x$ we have
$$
|f(x)-L| < epsilon
$$
but also, as $f$ decreases,
$$
|f(x) - L| geq |f(x_0)-L| > epsilon
$$
by choice of $epsilon$. Certainly $epsilon < |f(x)-L| < epsilon$ is a contradiction.
answered Jan 31 at 17:57
RandallRandall
10.7k11431
$endgroup$
(Assuming decreasing means monotonically so, not strictly so. I don't think this matters.)
Let $L = lim_{x to infty} f(x)$. Suppose to the contrary that there is an $x_0$ with $f(x_0) < L$. Since $f$ is decreasing, this means that $f(x) leq f(x_0) < L$ whenever $x geq x_0$. So, "after" $x_0$, $f(x)$ only moves further away from $L$ because it decreases. This will contradict the existence of the limit. This is the intuition.
Pick $epsilon = frac{L-f(x_0)}{3}$. Then there is an $M$ such that if $x geq M$ then $|f(x)-L| < epsilon$. Pick an $x$ greater than both $M$ and $x_0$. Then for this $x$ we have
$$
|f(x)-L| < epsilon
$$
but also, as $f$ decreases,
$$
|f(x) - L| geq |f(x_0)-L| > epsilon
$$
by choice of $epsilon$. Certainly $epsilon < |f(x)-L| < epsilon$ is a contradiction.
answered Jan 31 at 17:57
RandallRandall
10.7k11431
edited Jan 31 at 18:14
answered Jan 31 at 17:57
RandallRandall
10.7k11431
answered Jan 31 at 17:57
RandallRandall
10.7k11431
answered Jan 31 at 17:57
RandallRandall
10.7k11431
10.7k11431
add a comment |
add a comment |
$begingroup$
Using the limit definition:
Let $displaystyle L = lim_{xtoinfty} f(x)$. Choose $epsilon > 0$. Then there exists $x_0$ such that for all $x ge x_0$ we have $|f(x)-L|<epsilon$. That implies that $f(x) > L - epsilon$ for $x ge x_0$. For $x < x_0$ we have $f(x)ge f(x_0)ge L-epsilon$.
Then $f(x) ge L-epsilon$ for all $x$. As $epsilon >0$ is arbitrary we have $f(x) ge L$ for all $x$.
answered Jan 31 at 18:37
jjagmathjjagmath
3387
$endgroup$
add a comment |
$begingroup$
Using the limit definition:
Let $displaystyle L = lim_{xtoinfty} f(x)$. Choose $epsilon > 0$. Then there exists $x_0$ such that for all $x ge x_0$ we have $|f(x)-L|<epsilon$. That implies that $f(x) > L - epsilon$ for $x ge x_0$. For $x < x_0$ we have $f(x)ge f(x_0)ge L-epsilon$.
Then $f(x) ge L-epsilon$ for all $x$. As $epsilon >0$ is arbitrary we have $f(x) ge L$ for all $x$.
answered Jan 31 at 18:37
jjagmathjjagmath
3387
$endgroup$
add a comment |
$begingroup$
Using the limit definition:
Let $displaystyle L = lim_{xtoinfty} f(x)$. Choose $epsilon > 0$. Then there exists $x_0$ such that for all $x ge x_0$ we have $|f(x)-L|<epsilon$. That implies that $f(x) > L - epsilon$ for $x ge x_0$. For $x < x_0$ we have $f(x)ge f(x_0)ge L-epsilon$.
Then $f(x) ge L-epsilon$ for all $x$. As $epsilon >0$ is arbitrary we have $f(x) ge L$ for all $x$.
answered Jan 31 at 18:37
jjagmathjjagmath
3387
$endgroup$
Using the limit definition:
Let $displaystyle L = lim_{xtoinfty} f(x)$. Choose $epsilon > 0$. Then there exists $x_0$ such that for all $x ge x_0$ we have $|f(x)-L|<epsilon$. That implies that $f(x) > L - epsilon$ for $x ge x_0$. For $x < x_0$ we have $f(x)ge f(x_0)ge L-epsilon$.
Then $f(x) ge L-epsilon$ for all $x$. As $epsilon >0$ is arbitrary we have $f(x) ge L$ for all $x$.
answered Jan 31 at 18:37
jjagmathjjagmath
3387
answered Jan 31 at 18:37
jjagmathjjagmath
3387
answered Jan 31 at 18:37
jjagmathjjagmath
3387
answered Jan 31 at 18:37
jjagmathjjagmath
3387
3387
add a comment |
add a comment |
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$begingroup$
Do you want to assume the limit exists?
$endgroup$
– Randall
Jan 31 at 17:37
$begingroup$
Yes, of course, I forgot to say that.
$endgroup$
– JustAnAmateur
Jan 31 at 17:38
$begingroup$
Thank you guys for your input! I added the constraint that the limit exists.
$endgroup$
– JustAnAmateur
Jan 31 at 17:39
$begingroup$
Since the function is decreasing, then yes, the statement is true.
$endgroup$
– Clayton
Jan 31 at 17:40
1
$begingroup$
Prove it by contradiction.
$endgroup$
– Randall
Jan 31 at 17:40