Proving that $lim_{n to infty}t_n = s$
up vote
0
down vote
favorite
I'm stuck on the following problem:
$s_n$ and $t_n$ are sequences, such that $s_n=t_n$ except for finitely many values of $n$. Explain why if $lim_{n rightarrow infty} s_n = s$, then also $lim_{n rightarrow infty}t_n = s$, using the definition of limit.
Actually, my problem is that I understand that thing, but I can't come up with the idea of using the definition (with $epsilon$ and $N$) to prove it.
Thank you in advance for any help!
real-analysis
edited Oct 26 at 15:51
Bungo
13.6k22147
asked Oct 26 at 15:18
Sergey Malinov
103
add a comment |
up vote
0
down vote
favorite
I'm stuck on the following problem:
$s_n$ and $t_n$ are sequences, such that $s_n=t_n$ except for finitely many values of $n$. Explain why if $lim_{n rightarrow infty} s_n = s$, then also $lim_{n rightarrow infty}t_n = s$, using the definition of limit.
Actually, my problem is that I understand that thing, but I can't come up with the idea of using the definition (with $epsilon$ and $N$) to prove it.
Thank you in advance for any help!
real-analysis
edited Oct 26 at 15:51
Bungo
13.6k22147
asked Oct 26 at 15:18
Sergey Malinov
103
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm stuck on the following problem:
$s_n$ and $t_n$ are sequences, such that $s_n=t_n$ except for finitely many values of $n$. Explain why if $lim_{n rightarrow infty} s_n = s$, then also $lim_{n rightarrow infty}t_n = s$, using the definition of limit.
Actually, my problem is that I understand that thing, but I can't come up with the idea of using the definition (with $epsilon$ and $N$) to prove it.
Thank you in advance for any help!
real-analysis
edited Oct 26 at 15:51
Bungo
13.6k22147
asked Oct 26 at 15:18
Sergey Malinov
103
I'm stuck on the following problem:
$s_n$ and $t_n$ are sequences, such that $s_n=t_n$ except for finitely many values of $n$. Explain why if $lim_{n rightarrow infty} s_n = s$, then also $lim_{n rightarrow infty}t_n = s$, using the definition of limit.
Actually, my problem is that I understand that thing, but I can't come up with the idea of using the definition (with $epsilon$ and $N$) to prove it.
Thank you in advance for any help!
real-analysis
real-analysis
edited Oct 26 at 15:51
Bungo
13.6k22147
asked Oct 26 at 15:18
Sergey Malinov
103
edited Oct 26 at 15:51
Bungo
13.6k22147
asked Oct 26 at 15:18
Sergey Malinov
103
edited Oct 26 at 15:51
Bungo
13.6k22147
edited Oct 26 at 15:51
Bungo
13.6k22147
edited Oct 26 at 15:51
Bungo
13.6k22147
13.6k22147
asked Oct 26 at 15:18
Sergey Malinov
103
asked Oct 26 at 15:18
Sergey Malinov
103
asked Oct 26 at 15:18
Sergey Malinov
103
103
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Since you know that $lim _{nto infty } s_n =s$, given a positive $epsilon$ you can find an $N$ such that $$nge N implies |s_n - s | < epsilon$$
On the other hand you can find an $M$ such that $$ n>M implies s_n=t_n$$
Let $K= max {N,M }$ and see what happens if $n>K$
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
1
down vote
HINT
Since $s_n=t_n$ except for finitely many values of $n$ a maximum value $bar n$ exixts such that $s_{bar n}neq t_{bar n}$.
Then refer to the definition of limit.
answered Oct 26 at 15:39
gimusi
88k74393
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Since you know that $lim _{nto infty } s_n =s$, given a positive $epsilon$ you can find an $N$ such that $$nge N implies |s_n - s | < epsilon$$
On the other hand you can find an $M$ such that $$ n>M implies s_n=t_n$$
Let $K= max {N,M }$ and see what happens if $n>K$
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
accepted
Since you know that $lim _{nto infty } s_n =s$, given a positive $epsilon$ you can find an $N$ such that $$nge N implies |s_n - s | < epsilon$$
On the other hand you can find an $M$ such that $$ n>M implies s_n=t_n$$
Let $K= max {N,M }$ and see what happens if $n>K$
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since you know that $lim _{nto infty } s_n =s$, given a positive $epsilon$ you can find an $N$ such that $$nge N implies |s_n - s | < epsilon$$
On the other hand you can find an $M$ such that $$ n>M implies s_n=t_n$$
Let $K= max {N,M }$ and see what happens if $n>K$
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
Since you know that $lim _{nto infty } s_n =s$, given a positive $epsilon$ you can find an $N$ such that $$nge N implies |s_n - s | < epsilon$$
On the other hand you can find an $M$ such that $$ n>M implies s_n=t_n$$
Let $K= max {N,M }$ and see what happens if $n>K$
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
edited Nov 19 at 10:24
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
answered Oct 26 at 15:47
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
up vote
1
down vote
HINT
Since $s_n=t_n$ except for finitely many values of $n$ a maximum value $bar n$ exixts such that $s_{bar n}neq t_{bar n}$.
Then refer to the definition of limit.
answered Oct 26 at 15:39
gimusi
88k74393
add a comment |
up vote
1
down vote
HINT
Since $s_n=t_n$ except for finitely many values of $n$ a maximum value $bar n$ exixts such that $s_{bar n}neq t_{bar n}$.
Then refer to the definition of limit.
answered Oct 26 at 15:39
gimusi
88k74393
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT
Since $s_n=t_n$ except for finitely many values of $n$ a maximum value $bar n$ exixts such that $s_{bar n}neq t_{bar n}$.
Then refer to the definition of limit.
answered Oct 26 at 15:39
gimusi
88k74393
HINT
Since $s_n=t_n$ except for finitely many values of $n$ a maximum value $bar n$ exixts such that $s_{bar n}neq t_{bar n}$.
Then refer to the definition of limit.
answered Oct 26 at 15:39
gimusi
88k74393
answered Oct 26 at 15:39
gimusi
88k74393
answered Oct 26 at 15:39
gimusi
88k74393
answered Oct 26 at 15:39
gimusi
88k74393
88k74393
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2972280%2fproving-that-lim-n-to-inftyt-n-s%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
