if a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that $exists M >0$ , $|f(x)| leq M$...
$begingroup$
The question is given below:
Let $f$ be a continuous function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that there exists $M >0$ such that $|f(x)| leq M$ for all $x in [a, +infty).$
My Thoughts
1-But I wonder how can I prove this, will I use a proof similar to the comparison test?
2- Will the proof include the use this problem:
Let $f$ be an increasing function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that $f(x) leq L$ for all $x in [a, +infty).$and that $f$ is bounded on $[a, +infty).$
Any help will be appreciated.
real-analysis calculus improper-integrals
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
$endgroup$
add a comment |
$begingroup$
The question is given below:
Let $f$ be a continuous function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that there exists $M >0$ such that $|f(x)| leq M$ for all $x in [a, +infty).$
My Thoughts
1-But I wonder how can I prove this, will I use a proof similar to the comparison test?
2- Will the proof include the use this problem:
Let $f$ be an increasing function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that $f(x) leq L$ for all $x in [a, +infty).$and that $f$ is bounded on $[a, +infty).$
Any help will be appreciated.
real-analysis calculus improper-integrals
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
$endgroup$
add a comment |
$begingroup$
The question is given below:
Let $f$ be a continuous function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that there exists $M >0$ such that $|f(x)| leq M$ for all $x in [a, +infty).$
My Thoughts
1-But I wonder how can I prove this, will I use a proof similar to the comparison test?
2- Will the proof include the use this problem:
Let $f$ be an increasing function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that $f(x) leq L$ for all $x in [a, +infty).$and that $f$ is bounded on $[a, +infty).$
Any help will be appreciated.
real-analysis calculus improper-integrals
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
$endgroup$
The question is given below:
Let $f$ be a continuous function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that there exists $M >0$ such that $|f(x)| leq M$ for all $x in [a, +infty).$
My Thoughts
1-But I wonder how can I prove this, will I use a proof similar to the comparison test?
2- Will the proof include the use this problem:
Let $f$ be an increasing function for $x geq a,$ and suppose that a finite limit $L = lim_{x rightarrow infty} f(x) $ exists. Prove that $f(x) leq L$ for all $x in [a, +infty).$and that $f$ is bounded on $[a, +infty).$
Any help will be appreciated.
real-analysis calculus improper-integrals
real-analysis calculus improper-integrals
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
edited Dec 7 '18 at 5:24
hopefully
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
asked Dec 7 '18 at 5:17
hopefullyhopefully
290113
290113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By definition of limit, we know there exist a $Xin [a,+infty)$, such that
$$|f(x)-L|<1,,,,x>X$$
which implies that
$$|f(x)|<1+|L|,,,,x>X$$
If $X=a$, let $M=max{(1+|L|,|f(a)|)}$, then we've done.
If $X>a$, we know there exist $N>0$ such that
$$|f(x)|<N,,,xin[a,X]$$
by the continuity of $f$ on $[a,X]$. Let $M=max{(1+|L|,N)}$, we've done.
answered Dec 7 '18 at 5:28
LauLau
527315
$endgroup$
$begingroup$
Could you please clarify from where your forth line come ?
$endgroup$
– hopefully
Dec 7 '18 at 7:55
1
$begingroup$
$|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<max(|L-1|,|L+1|) leq 1+|L|$$
$endgroup$
– Lau
Dec 7 '18 at 7:58
$begingroup$
is this a well known fact?
$endgroup$
– hopefully
Dec 7 '18 at 8:22
$begingroup$
I think so .....
$endgroup$
– Lau
Dec 7 '18 at 8:24
1
$begingroup$
@hopefully edited
$endgroup$
– Lau
Dec 7 '18 at 9:05
|
show 1 more comment
$begingroup$
We know that for all $epsilon > 0$, there is an $N>0$ such that $x>N$, then $|f(x)-L| < epsilon$.
In particular, there is an $N_1>0$, such that $x > N$, then $|f(x)-L| < 1$. then we have $|f(x) | < |L|+1$.
Let $W = max_{x in [a, N]} |f(x)|.$
I'm leaving the last step for you to use these to construct an upper bound for $|f(x)|$.
Remark: We can't assume that $f$ is monotonic.
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
Could you provide some details about this last step please?
$endgroup$
– hopefully
Dec 7 '18 at 6:03
1
$begingroup$
I think $L+1$ is not necessary a positive number.
$endgroup$
– Lau
Dec 7 '18 at 6:08
$begingroup$
true, thanks for the feedback.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:24
$begingroup$
@hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:28
$begingroup$
Could you please clarify from where your third line come?
$endgroup$
– hopefully
Dec 7 '18 at 7:57
|
show 3 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By definition of limit, we know there exist a $Xin [a,+infty)$, such that
$$|f(x)-L|<1,,,,x>X$$
which implies that
$$|f(x)|<1+|L|,,,,x>X$$
If $X=a$, let $M=max{(1+|L|,|f(a)|)}$, then we've done.
If $X>a$, we know there exist $N>0$ such that
$$|f(x)|<N,,,xin[a,X]$$
by the continuity of $f$ on $[a,X]$. Let $M=max{(1+|L|,N)}$, we've done.
answered Dec 7 '18 at 5:28
LauLau
527315
$endgroup$
$begingroup$
Could you please clarify from where your forth line come ?
$endgroup$
– hopefully
Dec 7 '18 at 7:55
1
$begingroup$
$|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<max(|L-1|,|L+1|) leq 1+|L|$$
$endgroup$
– Lau
Dec 7 '18 at 7:58
$begingroup$
is this a well known fact?
$endgroup$
– hopefully
Dec 7 '18 at 8:22
$begingroup$
I think so .....
$endgroup$
– Lau
Dec 7 '18 at 8:24
1
$begingroup$
@hopefully edited
$endgroup$
– Lau
Dec 7 '18 at 9:05
|
show 1 more comment
$begingroup$
By definition of limit, we know there exist a $Xin [a,+infty)$, such that
$$|f(x)-L|<1,,,,x>X$$
which implies that
$$|f(x)|<1+|L|,,,,x>X$$
If $X=a$, let $M=max{(1+|L|,|f(a)|)}$, then we've done.
If $X>a$, we know there exist $N>0$ such that
$$|f(x)|<N,,,xin[a,X]$$
by the continuity of $f$ on $[a,X]$. Let $M=max{(1+|L|,N)}$, we've done.
answered Dec 7 '18 at 5:28
LauLau
527315
$endgroup$
$begingroup$
Could you please clarify from where your forth line come ?
$endgroup$
– hopefully
Dec 7 '18 at 7:55
1
$begingroup$
$|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<max(|L-1|,|L+1|) leq 1+|L|$$
$endgroup$
– Lau
Dec 7 '18 at 7:58
$begingroup$
is this a well known fact?
$endgroup$
– hopefully
Dec 7 '18 at 8:22
$begingroup$
I think so .....
$endgroup$
– Lau
Dec 7 '18 at 8:24
1
$begingroup$
@hopefully edited
$endgroup$
– Lau
Dec 7 '18 at 9:05
|
show 1 more comment
$begingroup$
By definition of limit, we know there exist a $Xin [a,+infty)$, such that
$$|f(x)-L|<1,,,,x>X$$
which implies that
$$|f(x)|<1+|L|,,,,x>X$$
If $X=a$, let $M=max{(1+|L|,|f(a)|)}$, then we've done.
If $X>a$, we know there exist $N>0$ such that
$$|f(x)|<N,,,xin[a,X]$$
by the continuity of $f$ on $[a,X]$. Let $M=max{(1+|L|,N)}$, we've done.
answered Dec 7 '18 at 5:28
LauLau
527315
$endgroup$
By definition of limit, we know there exist a $Xin [a,+infty)$, such that
$$|f(x)-L|<1,,,,x>X$$
which implies that
$$|f(x)|<1+|L|,,,,x>X$$
If $X=a$, let $M=max{(1+|L|,|f(a)|)}$, then we've done.
If $X>a$, we know there exist $N>0$ such that
$$|f(x)|<N,,,xin[a,X]$$
by the continuity of $f$ on $[a,X]$. Let $M=max{(1+|L|,N)}$, we've done.
answered Dec 7 '18 at 5:28
LauLau
527315
edited Dec 7 '18 at 9:05
answered Dec 7 '18 at 5:28
LauLau
527315
answered Dec 7 '18 at 5:28
LauLau
527315
answered Dec 7 '18 at 5:28
LauLau
527315
527315
$begingroup$
Could you please clarify from where your forth line come ?
$endgroup$
– hopefully
Dec 7 '18 at 7:55
1
$begingroup$
$|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<max(|L-1|,|L+1|) leq 1+|L|$$
$endgroup$
– Lau
Dec 7 '18 at 7:58
$begingroup$
is this a well known fact?
$endgroup$
– hopefully
Dec 7 '18 at 8:22
$begingroup$
I think so .....
$endgroup$
– Lau
Dec 7 '18 at 8:24
1
$begingroup$
@hopefully edited
$endgroup$
– Lau
Dec 7 '18 at 9:05
|
show 1 more comment
$begingroup$
Could you please clarify from where your forth line come ?
$endgroup$
– hopefully
Dec 7 '18 at 7:55
1
$begingroup$
$|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<max(|L-1|,|L+1|) leq 1+|L|$$
$endgroup$
– Lau
Dec 7 '18 at 7:58
$begingroup$
is this a well known fact?
$endgroup$
– hopefully
Dec 7 '18 at 8:22
$begingroup$
I think so .....
$endgroup$
– Lau
Dec 7 '18 at 8:24
1
$begingroup$
@hopefully edited
$endgroup$
– Lau
Dec 7 '18 at 9:05
$begingroup$
Could you please clarify from where your forth line come ?
$endgroup$
– hopefully
Dec 7 '18 at 7:55
$begingroup$
Could you please clarify from where your forth line come ?
$endgroup$
– hopefully
Dec 7 '18 at 7:55
1
1
$begingroup$
$|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<max(|L-1|,|L+1|) leq 1+|L|$$
$endgroup$
– Lau
Dec 7 '18 at 7:58
$begingroup$
$|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<max(|L-1|,|L+1|) leq 1+|L|$$
$endgroup$
– Lau
Dec 7 '18 at 7:58
$begingroup$
is this a well known fact?
$endgroup$
– hopefully
Dec 7 '18 at 8:22
$begingroup$
is this a well known fact?
$endgroup$
– hopefully
Dec 7 '18 at 8:22
$begingroup$
I think so .....
$endgroup$
– Lau
Dec 7 '18 at 8:24
$begingroup$
I think so .....
$endgroup$
– Lau
Dec 7 '18 at 8:24
1
1
$begingroup$
@hopefully edited
$endgroup$
– Lau
Dec 7 '18 at 9:05
$begingroup$
@hopefully edited
$endgroup$
– Lau
Dec 7 '18 at 9:05
|
show 1 more comment
$begingroup$
We know that for all $epsilon > 0$, there is an $N>0$ such that $x>N$, then $|f(x)-L| < epsilon$.
In particular, there is an $N_1>0$, such that $x > N$, then $|f(x)-L| < 1$. then we have $|f(x) | < |L|+1$.
Let $W = max_{x in [a, N]} |f(x)|.$
I'm leaving the last step for you to use these to construct an upper bound for $|f(x)|$.
Remark: We can't assume that $f$ is monotonic.
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
Could you provide some details about this last step please?
$endgroup$
– hopefully
Dec 7 '18 at 6:03
1
$begingroup$
I think $L+1$ is not necessary a positive number.
$endgroup$
– Lau
Dec 7 '18 at 6:08
$begingroup$
true, thanks for the feedback.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:24
$begingroup$
@hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:28
$begingroup$
Could you please clarify from where your third line come?
$endgroup$
– hopefully
Dec 7 '18 at 7:57
|
show 3 more comments
$begingroup$
We know that for all $epsilon > 0$, there is an $N>0$ such that $x>N$, then $|f(x)-L| < epsilon$.
In particular, there is an $N_1>0$, such that $x > N$, then $|f(x)-L| < 1$. then we have $|f(x) | < |L|+1$.
Let $W = max_{x in [a, N]} |f(x)|.$
I'm leaving the last step for you to use these to construct an upper bound for $|f(x)|$.
Remark: We can't assume that $f$ is monotonic.
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
Could you provide some details about this last step please?
$endgroup$
– hopefully
Dec 7 '18 at 6:03
1
$begingroup$
I think $L+1$ is not necessary a positive number.
$endgroup$
– Lau
Dec 7 '18 at 6:08
$begingroup$
true, thanks for the feedback.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:24
$begingroup$
@hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:28
$begingroup$
Could you please clarify from where your third line come?
$endgroup$
– hopefully
Dec 7 '18 at 7:57
|
show 3 more comments
$begingroup$
We know that for all $epsilon > 0$, there is an $N>0$ such that $x>N$, then $|f(x)-L| < epsilon$.
In particular, there is an $N_1>0$, such that $x > N$, then $|f(x)-L| < 1$. then we have $|f(x) | < |L|+1$.
Let $W = max_{x in [a, N]} |f(x)|.$
I'm leaving the last step for you to use these to construct an upper bound for $|f(x)|$.
Remark: We can't assume that $f$ is monotonic.
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
We know that for all $epsilon > 0$, there is an $N>0$ such that $x>N$, then $|f(x)-L| < epsilon$.
In particular, there is an $N_1>0$, such that $x > N$, then $|f(x)-L| < 1$. then we have $|f(x) | < |L|+1$.
Let $W = max_{x in [a, N]} |f(x)|.$
I'm leaving the last step for you to use these to construct an upper bound for $|f(x)|$.
Remark: We can't assume that $f$ is monotonic.
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
edited Dec 7 '18 at 6:24
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 7 '18 at 5:23
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
$begingroup$
Could you provide some details about this last step please?
$endgroup$
– hopefully
Dec 7 '18 at 6:03
1
$begingroup$
I think $L+1$ is not necessary a positive number.
$endgroup$
– Lau
Dec 7 '18 at 6:08
$begingroup$
true, thanks for the feedback.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:24
$begingroup$
@hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:28
$begingroup$
Could you please clarify from where your third line come?
$endgroup$
– hopefully
Dec 7 '18 at 7:57
|
show 3 more comments
$begingroup$
Could you provide some details about this last step please?
$endgroup$
– hopefully
Dec 7 '18 at 6:03
1
$begingroup$
I think $L+1$ is not necessary a positive number.
$endgroup$
– Lau
Dec 7 '18 at 6:08
$begingroup$
true, thanks for the feedback.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:24
$begingroup$
@hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:28
$begingroup$
Could you please clarify from where your third line come?
$endgroup$
– hopefully
Dec 7 '18 at 7:57
$begingroup$
Could you provide some details about this last step please?
$endgroup$
– hopefully
Dec 7 '18 at 6:03
$begingroup$
Could you provide some details about this last step please?
$endgroup$
– hopefully
Dec 7 '18 at 6:03
1
1
$begingroup$
I think $L+1$ is not necessary a positive number.
$endgroup$
– Lau
Dec 7 '18 at 6:08
$begingroup$
I think $L+1$ is not necessary a positive number.
$endgroup$
– Lau
Dec 7 '18 at 6:08
$begingroup$
true, thanks for the feedback.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:24
$begingroup$
true, thanks for the feedback.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:24
$begingroup$
@hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:28
$begingroup$
@hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 6:28
$begingroup$
Could you please clarify from where your third line come?
$endgroup$
– hopefully
Dec 7 '18 at 7:57
$begingroup$
Could you please clarify from where your third line come?
$endgroup$
– hopefully
Dec 7 '18 at 7:57
|
show 3 more comments
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