Mixing
Prove that $lim_{n to infty} (frac{n^2-1}{2n^2+3})=frac{1}{2}$ and $lim_{ntoinfty} frac{1}{ln(n+1)}=0$
$begingroup$
Use the definition of the limit of a sequence to prove that $lim_{n to infty} (frac{n^2-1}{2n^2+3})=frac{1}{2}$.
We have
$$begin{align}
left|frac{n^2-1}{2n^2+3}-frac{1}{2}right| & =left|frac{2n^2-2-2n^2-3}{2(2n^2+3)}right| \
&= left|frac{-5}{2(2n^2+3)}right|\
&= frac{5}{2(2n^2+3)} \
&<frac{5}{4n^2},
end{align}$$
$$frac{5}{4n^2}<epsilon iff frac{1}{n^2}<frac{4 epsilon}{5} iff n>sqrt{frac{5}{4epsilon}}$$
We choose $n_0=left[sqrt{frac{5}{4epsilon}} right]+1$, Then $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$.
Let $(x_n)=frac{1}{ln(n+1)}$ for $n in mathbb{N}$.
a) Use the definition of the limit to show that $lim(x_n)=0$.
$|frac{1}{ln(n+1)}-0|=frac{1}{ln(n+1)}<epsilon Leftrightarrow ln(n+1) > epsilon Leftrightarrow n> e^{epsilon} -1$
We choose $n_0=left[ e^epsilon -1 right]+1$, Then $lim(x_n)=0$.
b) Find specific value of $n_0 (epsilon)$ as required in definition of limit for $epsilon=frac{1}{2}$.
$n_0=left[sqrt{e}-1right]+1$
Is that true, please?
real-analysis sequences-and-series limits
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
asked Feb 3 at 1:14
DimaDima
873616
$endgroup$
add a comment |
$begingroup$
Use the definition of the limit of a sequence to prove that $lim_{n to infty} (frac{n^2-1}{2n^2+3})=frac{1}{2}$.
We have
$$begin{align}
left|frac{n^2-1}{2n^2+3}-frac{1}{2}right| & =left|frac{2n^2-2-2n^2-3}{2(2n^2+3)}right| \
&= left|frac{-5}{2(2n^2+3)}right|\
&= frac{5}{2(2n^2+3)} \
&<frac{5}{4n^2},
end{align}$$
$$frac{5}{4n^2}<epsilon iff frac{1}{n^2}<frac{4 epsilon}{5} iff n>sqrt{frac{5}{4epsilon}}$$
We choose $n_0=left[sqrt{frac{5}{4epsilon}} right]+1$, Then $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$.
Let $(x_n)=frac{1}{ln(n+1)}$ for $n in mathbb{N}$.
a) Use the definition of the limit to show that $lim(x_n)=0$.
$|frac{1}{ln(n+1)}-0|=frac{1}{ln(n+1)}<epsilon Leftrightarrow ln(n+1) > epsilon Leftrightarrow n> e^{epsilon} -1$
We choose $n_0=left[ e^epsilon -1 right]+1$, Then $lim(x_n)=0$.
b) Find specific value of $n_0 (epsilon)$ as required in definition of limit for $epsilon=frac{1}{2}$.
$n_0=left[sqrt{e}-1right]+1$
Is that true, please?
real-analysis sequences-and-series limits
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
asked Feb 3 at 1:14
DimaDima
873616
$endgroup$
1
$begingroup$
In your second exercise, you should have $ln(n+1)>1/epsilon$.
$endgroup$
– John Wayland Bales
Feb 3 at 1:25
1
$begingroup$
You are only showing the work in finding an $n_0$. The actual proof should look like "Let $epsilon > 0$, and let $n_0 =$ ____. Then if $n>n_0$ .... work ... $|a_n - L| < epsilon$. QED"
$endgroup$
– David Peterson
Feb 3 at 1:26
$begingroup$
@JohnWaylandBales You are right. Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
$begingroup$
@DavidPeterson Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
add a comment |
$begingroup$
Use the definition of the limit of a sequence to prove that $lim_{n to infty} (frac{n^2-1}{2n^2+3})=frac{1}{2}$.
We have
$$begin{align}
left|frac{n^2-1}{2n^2+3}-frac{1}{2}right| & =left|frac{2n^2-2-2n^2-3}{2(2n^2+3)}right| \
&= left|frac{-5}{2(2n^2+3)}right|\
&= frac{5}{2(2n^2+3)} \
&<frac{5}{4n^2},
end{align}$$
$$frac{5}{4n^2}<epsilon iff frac{1}{n^2}<frac{4 epsilon}{5} iff n>sqrt{frac{5}{4epsilon}}$$
We choose $n_0=left[sqrt{frac{5}{4epsilon}} right]+1$, Then $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$.
Let $(x_n)=frac{1}{ln(n+1)}$ for $n in mathbb{N}$.
a) Use the definition of the limit to show that $lim(x_n)=0$.
$|frac{1}{ln(n+1)}-0|=frac{1}{ln(n+1)}<epsilon Leftrightarrow ln(n+1) > epsilon Leftrightarrow n> e^{epsilon} -1$
We choose $n_0=left[ e^epsilon -1 right]+1$, Then $lim(x_n)=0$.
b) Find specific value of $n_0 (epsilon)$ as required in definition of limit for $epsilon=frac{1}{2}$.
$n_0=left[sqrt{e}-1right]+1$
Is that true, please?
real-analysis sequences-and-series limits
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
asked Feb 3 at 1:14
DimaDima
873616
$endgroup$
Use the definition of the limit of a sequence to prove that $lim_{n to infty} (frac{n^2-1}{2n^2+3})=frac{1}{2}$.
We have
$$begin{align}
left|frac{n^2-1}{2n^2+3}-frac{1}{2}right| & =left|frac{2n^2-2-2n^2-3}{2(2n^2+3)}right| \
&= left|frac{-5}{2(2n^2+3)}right|\
&= frac{5}{2(2n^2+3)} \
&<frac{5}{4n^2},
end{align}$$
$$frac{5}{4n^2}<epsilon iff frac{1}{n^2}<frac{4 epsilon}{5} iff n>sqrt{frac{5}{4epsilon}}$$
We choose $n_0=left[sqrt{frac{5}{4epsilon}} right]+1$, Then $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$.
Let $(x_n)=frac{1}{ln(n+1)}$ for $n in mathbb{N}$.
a) Use the definition of the limit to show that $lim(x_n)=0$.
$|frac{1}{ln(n+1)}-0|=frac{1}{ln(n+1)}<epsilon Leftrightarrow ln(n+1) > epsilon Leftrightarrow n> e^{epsilon} -1$
We choose $n_0=left[ e^epsilon -1 right]+1$, Then $lim(x_n)=0$.
b) Find specific value of $n_0 (epsilon)$ as required in definition of limit for $epsilon=frac{1}{2}$.
$n_0=left[sqrt{e}-1right]+1$
Is that true, please?
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
asked Feb 3 at 1:14
DimaDima
873616
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
asked Feb 3 at 1:14
DimaDima
873616
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
17.7k42881
asked Feb 3 at 1:14
DimaDima
873616
asked Feb 3 at 1:14
DimaDima
873616
asked Feb 3 at 1:14
DimaDima
873616
873616
1
$begingroup$
In your second exercise, you should have $ln(n+1)>1/epsilon$.
$endgroup$
– John Wayland Bales
Feb 3 at 1:25
1
$begingroup$
You are only showing the work in finding an $n_0$. The actual proof should look like "Let $epsilon > 0$, and let $n_0 =$ ____. Then if $n>n_0$ .... work ... $|a_n - L| < epsilon$. QED"
$endgroup$
– David Peterson
Feb 3 at 1:26
$begingroup$
@JohnWaylandBales You are right. Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
$begingroup$
@DavidPeterson Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
add a comment |
1
$begingroup$
In your second exercise, you should have $ln(n+1)>1/epsilon$.
$endgroup$
– John Wayland Bales
Feb 3 at 1:25
1
$begingroup$
You are only showing the work in finding an $n_0$. The actual proof should look like "Let $epsilon > 0$, and let $n_0 =$ ____. Then if $n>n_0$ .... work ... $|a_n - L| < epsilon$. QED"
$endgroup$
– David Peterson
Feb 3 at 1:26
$begingroup$
@JohnWaylandBales You are right. Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
$begingroup$
@DavidPeterson Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
1
1
$begingroup$
In your second exercise, you should have $ln(n+1)>1/epsilon$.
$endgroup$
– John Wayland Bales
Feb 3 at 1:25
$begingroup$
In your second exercise, you should have $ln(n+1)>1/epsilon$.
$endgroup$
– John Wayland Bales
Feb 3 at 1:25
1
1
$begingroup$
You are only showing the work in finding an $n_0$. The actual proof should look like "Let $epsilon > 0$, and let $n_0 =$ ____. Then if $n>n_0$ .... work ... $|a_n - L| < epsilon$. QED"
$endgroup$
– David Peterson
Feb 3 at 1:26
$begingroup$
You are only showing the work in finding an $n_0$. The actual proof should look like "Let $epsilon > 0$, and let $n_0 =$ ____. Then if $n>n_0$ .... work ... $|a_n - L| < epsilon$. QED"
$endgroup$
– David Peterson
Feb 3 at 1:26
$begingroup$
@JohnWaylandBales You are right. Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
$begingroup$
@JohnWaylandBales You are right. Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
$begingroup$
@DavidPeterson Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
$begingroup$
@DavidPeterson Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:
Prove: $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$
Proof: Let $epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $left|frac{n^2-1}{2n^2+3}-frac{1}{2}right|<epsilon$
Then proceed with the steps which you have given.
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:57
add a comment |
$begingroup$
For a) We have $lim_{nrightarrow infty} frac{n^2-1}{2n^2+3}=lim_{nrightarrow infty} frac{n^2(1-frac{1}{n^2})}{n^2(2+frac{3}{n^2})}= lim_{nrightarrow infty} frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}=frac{1}{2}$
For b) basically the same,meaning $ln$ is monotone so for $nrightarrow infty$ it follows that $ln(n)rightarrow infty$
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
$endgroup$
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
Your answer is correct. However it seems you may have overcomplicated it.
begin{align}
lim_{nto infty}left(frac{n^2-1}{2n^2+3}right) & = lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right)\
end{align}
Now use the fact that $lim_{nto infty}left(frac{1}{n^k}right)=0$, where $k$ is any positive integer.
Hence
begin{align}
lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right) & = lim_{nto infty}left(frac{1-0}{2+0}right)\
&= frac{1}{2}
end{align}
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
$endgroup$
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
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
$begingroup$
If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:
Prove: $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$
Proof: Let $epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $left|frac{n^2-1}{2n^2+3}-frac{1}{2}right|<epsilon$
Then proceed with the steps which you have given.
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:57
add a comment |
$begingroup$
If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:
Prove: $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$
Proof: Let $epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $left|frac{n^2-1}{2n^2+3}-frac{1}{2}right|<epsilon$
Then proceed with the steps which you have given.
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:57
add a comment |
$begingroup$
If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:
Prove: $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$
Proof: Let $epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $left|frac{n^2-1}{2n^2+3}-frac{1}{2}right|<epsilon$
Then proceed with the steps which you have given.
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
$endgroup$
If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:
Prove: $lim_{n to infty} left(frac{n^2-1}{2n^2+3}right)=frac{1}{2}$
Proof: Let $epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $left|frac{n^2-1}{2n^2+3}-frac{1}{2}right|<epsilon$
Then proceed with the steps which you have given.
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
answered Feb 3 at 1:32
John Wayland BalesJohn Wayland Bales
15.2k21238
15.2k21238
$begingroup$
Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:57
add a comment |
$begingroup$
Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:57
$begingroup$
Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:57
$begingroup$
Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:57
add a comment |
$begingroup$
For a) We have $lim_{nrightarrow infty} frac{n^2-1}{2n^2+3}=lim_{nrightarrow infty} frac{n^2(1-frac{1}{n^2})}{n^2(2+frac{3}{n^2})}= lim_{nrightarrow infty} frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}=frac{1}{2}$
For b) basically the same,meaning $ln$ is monotone so for $nrightarrow infty$ it follows that $ln(n)rightarrow infty$
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
$endgroup$
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
For a) We have $lim_{nrightarrow infty} frac{n^2-1}{2n^2+3}=lim_{nrightarrow infty} frac{n^2(1-frac{1}{n^2})}{n^2(2+frac{3}{n^2})}= lim_{nrightarrow infty} frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}=frac{1}{2}$
For b) basically the same,meaning $ln$ is monotone so for $nrightarrow infty$ it follows that $ln(n)rightarrow infty$
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
$endgroup$
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
For a) We have $lim_{nrightarrow infty} frac{n^2-1}{2n^2+3}=lim_{nrightarrow infty} frac{n^2(1-frac{1}{n^2})}{n^2(2+frac{3}{n^2})}= lim_{nrightarrow infty} frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}=frac{1}{2}$
For b) basically the same,meaning $ln$ is monotone so for $nrightarrow infty$ it follows that $ln(n)rightarrow infty$
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
$endgroup$
For a) We have $lim_{nrightarrow infty} frac{n^2-1}{2n^2+3}=lim_{nrightarrow infty} frac{n^2(1-frac{1}{n^2})}{n^2(2+frac{3}{n^2})}= lim_{nrightarrow infty} frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}=frac{1}{2}$
For b) basically the same,meaning $ln$ is monotone so for $nrightarrow infty$ it follows that $ln(n)rightarrow infty$
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
edited Feb 3 at 1:36
Hanul Jeon
17.7k42881
17.7k42881
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
answered Feb 3 at 1:22
babemcnuggetsbabemcnuggets
116110
116110
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
Your answer is correct. However it seems you may have overcomplicated it.
begin{align}
lim_{nto infty}left(frac{n^2-1}{2n^2+3}right) & = lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right)\
end{align}
Now use the fact that $lim_{nto infty}left(frac{1}{n^k}right)=0$, where $k$ is any positive integer.
Hence
begin{align}
lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right) & = lim_{nto infty}left(frac{1-0}{2+0}right)\
&= frac{1}{2}
end{align}
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
$endgroup$
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
Your answer is correct. However it seems you may have overcomplicated it.
begin{align}
lim_{nto infty}left(frac{n^2-1}{2n^2+3}right) & = lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right)\
end{align}
Now use the fact that $lim_{nto infty}left(frac{1}{n^k}right)=0$, where $k$ is any positive integer.
Hence
begin{align}
lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right) & = lim_{nto infty}left(frac{1-0}{2+0}right)\
&= frac{1}{2}
end{align}
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
$endgroup$
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
Your answer is correct. However it seems you may have overcomplicated it.
begin{align}
lim_{nto infty}left(frac{n^2-1}{2n^2+3}right) & = lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right)\
end{align}
Now use the fact that $lim_{nto infty}left(frac{1}{n^k}right)=0$, where $k$ is any positive integer.
Hence
begin{align}
lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right) & = lim_{nto infty}left(frac{1-0}{2+0}right)\
&= frac{1}{2}
end{align}
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
$endgroup$
Your answer is correct. However it seems you may have overcomplicated it.
begin{align}
lim_{nto infty}left(frac{n^2-1}{2n^2+3}right) & = lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right)\
end{align}
Now use the fact that $lim_{nto infty}left(frac{1}{n^k}right)=0$, where $k$ is any positive integer.
Hence
begin{align}
lim_{nto infty}left(frac{1-frac{1}{n^2}}{2+frac{3}{n^2}}right) & = lim_{nto infty}left(frac{1-0}{2+0}right)\
&= frac{1}{2}
end{align}
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
edited Feb 3 at 1:37
Hanul Jeon
17.7k42881
17.7k42881
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
answered Feb 3 at 1:23
LandurosLanduros
1,8871620
1,8871620
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
$begingroup$
Thanks a lot, sorry I forget to write "by the definition".
$endgroup$
– Dima
Feb 3 at 1:34
add a comment |
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$begingroup$
In your second exercise, you should have $ln(n+1)>1/epsilon$.
$endgroup$
– John Wayland Bales
Feb 3 at 1:25
1
$begingroup$
You are only showing the work in finding an $n_0$. The actual proof should look like "Let $epsilon > 0$, and let $n_0 =$ ____. Then if $n>n_0$ .... work ... $|a_n - L| < epsilon$. QED"
$endgroup$
– David Peterson
Feb 3 at 1:26
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@JohnWaylandBales You are right. Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35
$begingroup$
@DavidPeterson Thank you so much.
$endgroup$
– Dima
Feb 3 at 1:35