Another Epsilon-N Limit Proof Question
How to prove the limit of the following sequence using epsilon-N argument.
$$a_n=frac{3n^2+2n+1}{2n^2+n}$$
I took the limit to be $frac{3}{2}$ and proceeded with the argument,
$$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<epsilon$$
$$frac{n+2}{4n^2+2n}<epsilon$$
How do I complete the argument from this point?
calculus real-analysis sequences-and-series limits epsilon-delta
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
asked Mar 2 '14 at 18:58
user123652
65113
|
show 1 more comment
How to prove the limit of the following sequence using epsilon-N argument.
$$a_n=frac{3n^2+2n+1}{2n^2+n}$$
I took the limit to be $frac{3}{2}$ and proceeded with the argument,
$$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<epsilon$$
$$frac{n+2}{4n^2+2n}<epsilon$$
How do I complete the argument from this point?
calculus real-analysis sequences-and-series limits epsilon-delta
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
asked Mar 2 '14 at 18:58
user123652
65113
thank you, I'm new to these proofs should have noticed that.
– user123652
Mar 2 '14 at 19:03
@AlijahAhmed What you say is false. It's easy to check for $n=1$.
– Git Gud
Mar 2 '14 at 19:04
I believe you are arguing what you want to prove, in a way. Are you sure the $epsilon - N$ methodology is clear to you?
– user76568
Mar 2 '14 at 19:05
Sorry, completely messed up the factorisation (rather foolish error!)- the fraction $frac{n+2}{4n^2+2n}=frac{(n+2)}{2n(2n+1)}$, see @Dror's answer for a comprehensive answer.
– Alijah Ahmed
Mar 2 '14 at 19:27
@AlijahAhmed I wouldn't call my answer comprehensive... Unless you were being sarcastic..?
– user76568
Mar 2 '14 at 19:50
|
show 1 more comment
How to prove the limit of the following sequence using epsilon-N argument.
$$a_n=frac{3n^2+2n+1}{2n^2+n}$$
I took the limit to be $frac{3}{2}$ and proceeded with the argument,
$$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<epsilon$$
$$frac{n+2}{4n^2+2n}<epsilon$$
How do I complete the argument from this point?
calculus real-analysis sequences-and-series limits epsilon-delta
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
asked Mar 2 '14 at 18:58
user123652
65113
How to prove the limit of the following sequence using epsilon-N argument.
$$a_n=frac{3n^2+2n+1}{2n^2+n}$$
I took the limit to be $frac{3}{2}$ and proceeded with the argument,
$$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<epsilon$$
$$frac{n+2}{4n^2+2n}<epsilon$$
How do I complete the argument from this point?
calculus real-analysis sequences-and-series limits epsilon-delta
calculus real-analysis sequences-and-series limits epsilon-delta
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
asked Mar 2 '14 at 18:58
user123652
65113
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
asked Mar 2 '14 at 18:58
user123652
65113
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
edited Jan 2 '15 at 17:23
Git Gud
28.7k1049100
28.7k1049100
asked Mar 2 '14 at 18:58
user123652
65113
asked Mar 2 '14 at 18:58
user123652
65113
asked Mar 2 '14 at 18:58
user123652
65113
65113
thank you, I'm new to these proofs should have noticed that.
– user123652
Mar 2 '14 at 19:03
@AlijahAhmed What you say is false. It's easy to check for $n=1$.
– Git Gud
Mar 2 '14 at 19:04
I believe you are arguing what you want to prove, in a way. Are you sure the $epsilon - N$ methodology is clear to you?
– user76568
Mar 2 '14 at 19:05
Sorry, completely messed up the factorisation (rather foolish error!)- the fraction $frac{n+2}{4n^2+2n}=frac{(n+2)}{2n(2n+1)}$, see @Dror's answer for a comprehensive answer.
– Alijah Ahmed
Mar 2 '14 at 19:27
@AlijahAhmed I wouldn't call my answer comprehensive... Unless you were being sarcastic..?
– user76568
Mar 2 '14 at 19:50
|
show 1 more comment
thank you, I'm new to these proofs should have noticed that.
– user123652
Mar 2 '14 at 19:03
@AlijahAhmed What you say is false. It's easy to check for $n=1$.
– Git Gud
Mar 2 '14 at 19:04
I believe you are arguing what you want to prove, in a way. Are you sure the $epsilon - N$ methodology is clear to you?
– user76568
Mar 2 '14 at 19:05
Sorry, completely messed up the factorisation (rather foolish error!)- the fraction $frac{n+2}{4n^2+2n}=frac{(n+2)}{2n(2n+1)}$, see @Dror's answer for a comprehensive answer.
– Alijah Ahmed
Mar 2 '14 at 19:27
@AlijahAhmed I wouldn't call my answer comprehensive... Unless you were being sarcastic..?
– user76568
Mar 2 '14 at 19:50
thank you, I'm new to these proofs should have noticed that.
– user123652
Mar 2 '14 at 19:03
thank you, I'm new to these proofs should have noticed that.
– user123652
Mar 2 '14 at 19:03
@AlijahAhmed What you say is false. It's easy to check for $n=1$.
– Git Gud
Mar 2 '14 at 19:04
@AlijahAhmed What you say is false. It's easy to check for $n=1$.
– Git Gud
Mar 2 '14 at 19:04
I believe you are arguing what you want to prove, in a way. Are you sure the $epsilon - N$ methodology is clear to you?
– user76568
Mar 2 '14 at 19:05
I believe you are arguing what you want to prove, in a way. Are you sure the $epsilon - N$ methodology is clear to you?
– user76568
Mar 2 '14 at 19:05
Sorry, completely messed up the factorisation (rather foolish error!)- the fraction $frac{n+2}{4n^2+2n}=frac{(n+2)}{2n(2n+1)}$, see @Dror's answer for a comprehensive answer.
– Alijah Ahmed
Mar 2 '14 at 19:27
Sorry, completely messed up the factorisation (rather foolish error!)- the fraction $frac{n+2}{4n^2+2n}=frac{(n+2)}{2n(2n+1)}$, see @Dror's answer for a comprehensive answer.
– Alijah Ahmed
Mar 2 '14 at 19:27
@AlijahAhmed I wouldn't call my answer comprehensive... Unless you were being sarcastic..?
– user76568
Mar 2 '14 at 19:50
@AlijahAhmed I wouldn't call my answer comprehensive... Unless you were being sarcastic..?
– user76568
Mar 2 '14 at 19:50
|
show 1 more comment
2 Answers
2
active
oldest
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$$frac{n+2}{4n^2 + 2n} = frac{n+2}{2n(2n+1)} leq frac{n+2}{2n(n+2)} = frac{1}{2n}$$
Do you want to take it from here? Maybe you'd like to see a full $epsilon - N$ proof?
answered Mar 2 '14 at 19:22
user76568
3,5631129
add a comment |
When you get an expression of the form $f(n)<varepsilon$, what you'd like to do is rewrite the inequality in terms of $varepsilon$, that is, you'd like to write $g(varepsilon )<n$, for some $g$. Then you take $n$ a natural number that satisfies this inequality and you're done.
Often times $f(n)$ is not simple enough (sometimes even impossible) to do this, so you need to find another function $h$ for the same purpose (rewriting $h(n)<varepsilon$ with respect to $varepsilon$) with the additional properties that $f(n)<h(n)$ and $lim limits_{nto +infty}left(h(n)right)=0$.
It's easy to see that $dfrac{n+2}{4n^2+2n}<dfrac{n+2}{4n^2}$.
Let $h(n)=dfrac {n+2}{4n^2}$ and consider $h(n)<varepsilon$.
Now consider $h(n)=varepsilon$. This is equivalent to $4n^2varepsilon -n-2=0$ which in turn is equivalent to $n=dfrac{1pm sqrt{1+32varepsilon}}{8varepsilon}$, so the inequality $h(n)<varepsilon$ will be satisfied just as long as $n$ is a natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$.
Getting back to your proof, you wish to prove that
$$forall varepsilon >0exists Nin mathbb N forall nin mathbb Nleft(nge Nimplies left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilonright).$$
Take $varepsilon>0$ and let $N$ be any natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$. Take any natural number $n$ greater than $N$.
Your goal is to prove that $$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon,$$
which is equivalent to proving $$frac{n+2}{4n^2+2n}<epsilon.$$
You know that $N$ satisfies $fleft(Nright)<hleft(Nright)<varepsilon$, i.e., $N$ satisfies the inequality above. It's easy to prove that because $nge N$, then $n$ will satisfy $f(n)<fleft(Nright)$ and consequently it will satisfy the inequality $left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon$, as is needed.
Of course Dror's $h$ in his answer is much, much simpler, but my remarks are still compatible with his choice for $h$
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
add a comment |
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2 Answers
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$$frac{n+2}{4n^2 + 2n} = frac{n+2}{2n(2n+1)} leq frac{n+2}{2n(n+2)} = frac{1}{2n}$$
Do you want to take it from here? Maybe you'd like to see a full $epsilon - N$ proof?
answered Mar 2 '14 at 19:22
user76568
3,5631129
add a comment |
$$frac{n+2}{4n^2 + 2n} = frac{n+2}{2n(2n+1)} leq frac{n+2}{2n(n+2)} = frac{1}{2n}$$
Do you want to take it from here? Maybe you'd like to see a full $epsilon - N$ proof?
answered Mar 2 '14 at 19:22
user76568
3,5631129
add a comment |
$$frac{n+2}{4n^2 + 2n} = frac{n+2}{2n(2n+1)} leq frac{n+2}{2n(n+2)} = frac{1}{2n}$$
Do you want to take it from here? Maybe you'd like to see a full $epsilon - N$ proof?
answered Mar 2 '14 at 19:22
user76568
3,5631129
$$frac{n+2}{4n^2 + 2n} = frac{n+2}{2n(2n+1)} leq frac{n+2}{2n(n+2)} = frac{1}{2n}$$
Do you want to take it from here? Maybe you'd like to see a full $epsilon - N$ proof?
answered Mar 2 '14 at 19:22
user76568
3,5631129
edited Mar 2 '14 at 19:29
answered Mar 2 '14 at 19:22
user76568
3,5631129
answered Mar 2 '14 at 19:22
user76568
3,5631129
answered Mar 2 '14 at 19:22
user76568
3,5631129
3,5631129
add a comment |
add a comment |
When you get an expression of the form $f(n)<varepsilon$, what you'd like to do is rewrite the inequality in terms of $varepsilon$, that is, you'd like to write $g(varepsilon )<n$, for some $g$. Then you take $n$ a natural number that satisfies this inequality and you're done.
Often times $f(n)$ is not simple enough (sometimes even impossible) to do this, so you need to find another function $h$ for the same purpose (rewriting $h(n)<varepsilon$ with respect to $varepsilon$) with the additional properties that $f(n)<h(n)$ and $lim limits_{nto +infty}left(h(n)right)=0$.
It's easy to see that $dfrac{n+2}{4n^2+2n}<dfrac{n+2}{4n^2}$.
Let $h(n)=dfrac {n+2}{4n^2}$ and consider $h(n)<varepsilon$.
Now consider $h(n)=varepsilon$. This is equivalent to $4n^2varepsilon -n-2=0$ which in turn is equivalent to $n=dfrac{1pm sqrt{1+32varepsilon}}{8varepsilon}$, so the inequality $h(n)<varepsilon$ will be satisfied just as long as $n$ is a natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$.
Getting back to your proof, you wish to prove that
$$forall varepsilon >0exists Nin mathbb N forall nin mathbb Nleft(nge Nimplies left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilonright).$$
Take $varepsilon>0$ and let $N$ be any natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$. Take any natural number $n$ greater than $N$.
Your goal is to prove that $$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon,$$
which is equivalent to proving $$frac{n+2}{4n^2+2n}<epsilon.$$
You know that $N$ satisfies $fleft(Nright)<hleft(Nright)<varepsilon$, i.e., $N$ satisfies the inequality above. It's easy to prove that because $nge N$, then $n$ will satisfy $f(n)<fleft(Nright)$ and consequently it will satisfy the inequality $left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon$, as is needed.
Of course Dror's $h$ in his answer is much, much simpler, but my remarks are still compatible with his choice for $h$
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
add a comment |
When you get an expression of the form $f(n)<varepsilon$, what you'd like to do is rewrite the inequality in terms of $varepsilon$, that is, you'd like to write $g(varepsilon )<n$, for some $g$. Then you take $n$ a natural number that satisfies this inequality and you're done.
Often times $f(n)$ is not simple enough (sometimes even impossible) to do this, so you need to find another function $h$ for the same purpose (rewriting $h(n)<varepsilon$ with respect to $varepsilon$) with the additional properties that $f(n)<h(n)$ and $lim limits_{nto +infty}left(h(n)right)=0$.
It's easy to see that $dfrac{n+2}{4n^2+2n}<dfrac{n+2}{4n^2}$.
Let $h(n)=dfrac {n+2}{4n^2}$ and consider $h(n)<varepsilon$.
Now consider $h(n)=varepsilon$. This is equivalent to $4n^2varepsilon -n-2=0$ which in turn is equivalent to $n=dfrac{1pm sqrt{1+32varepsilon}}{8varepsilon}$, so the inequality $h(n)<varepsilon$ will be satisfied just as long as $n$ is a natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$.
Getting back to your proof, you wish to prove that
$$forall varepsilon >0exists Nin mathbb N forall nin mathbb Nleft(nge Nimplies left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilonright).$$
Take $varepsilon>0$ and let $N$ be any natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$. Take any natural number $n$ greater than $N$.
Your goal is to prove that $$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon,$$
which is equivalent to proving $$frac{n+2}{4n^2+2n}<epsilon.$$
You know that $N$ satisfies $fleft(Nright)<hleft(Nright)<varepsilon$, i.e., $N$ satisfies the inequality above. It's easy to prove that because $nge N$, then $n$ will satisfy $f(n)<fleft(Nright)$ and consequently it will satisfy the inequality $left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon$, as is needed.
Of course Dror's $h$ in his answer is much, much simpler, but my remarks are still compatible with his choice for $h$
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
add a comment |
When you get an expression of the form $f(n)<varepsilon$, what you'd like to do is rewrite the inequality in terms of $varepsilon$, that is, you'd like to write $g(varepsilon )<n$, for some $g$. Then you take $n$ a natural number that satisfies this inequality and you're done.
Often times $f(n)$ is not simple enough (sometimes even impossible) to do this, so you need to find another function $h$ for the same purpose (rewriting $h(n)<varepsilon$ with respect to $varepsilon$) with the additional properties that $f(n)<h(n)$ and $lim limits_{nto +infty}left(h(n)right)=0$.
It's easy to see that $dfrac{n+2}{4n^2+2n}<dfrac{n+2}{4n^2}$.
Let $h(n)=dfrac {n+2}{4n^2}$ and consider $h(n)<varepsilon$.
Now consider $h(n)=varepsilon$. This is equivalent to $4n^2varepsilon -n-2=0$ which in turn is equivalent to $n=dfrac{1pm sqrt{1+32varepsilon}}{8varepsilon}$, so the inequality $h(n)<varepsilon$ will be satisfied just as long as $n$ is a natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$.
Getting back to your proof, you wish to prove that
$$forall varepsilon >0exists Nin mathbb N forall nin mathbb Nleft(nge Nimplies left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilonright).$$
Take $varepsilon>0$ and let $N$ be any natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$. Take any natural number $n$ greater than $N$.
Your goal is to prove that $$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon,$$
which is equivalent to proving $$frac{n+2}{4n^2+2n}<epsilon.$$
You know that $N$ satisfies $fleft(Nright)<hleft(Nright)<varepsilon$, i.e., $N$ satisfies the inequality above. It's easy to prove that because $nge N$, then $n$ will satisfy $f(n)<fleft(Nright)$ and consequently it will satisfy the inequality $left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon$, as is needed.
Of course Dror's $h$ in his answer is much, much simpler, but my remarks are still compatible with his choice for $h$
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
When you get an expression of the form $f(n)<varepsilon$, what you'd like to do is rewrite the inequality in terms of $varepsilon$, that is, you'd like to write $g(varepsilon )<n$, for some $g$. Then you take $n$ a natural number that satisfies this inequality and you're done.
Often times $f(n)$ is not simple enough (sometimes even impossible) to do this, so you need to find another function $h$ for the same purpose (rewriting $h(n)<varepsilon$ with respect to $varepsilon$) with the additional properties that $f(n)<h(n)$ and $lim limits_{nto +infty}left(h(n)right)=0$.
It's easy to see that $dfrac{n+2}{4n^2+2n}<dfrac{n+2}{4n^2}$.
Let $h(n)=dfrac {n+2}{4n^2}$ and consider $h(n)<varepsilon$.
Now consider $h(n)=varepsilon$. This is equivalent to $4n^2varepsilon -n-2=0$ which in turn is equivalent to $n=dfrac{1pm sqrt{1+32varepsilon}}{8varepsilon}$, so the inequality $h(n)<varepsilon$ will be satisfied just as long as $n$ is a natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$.
Getting back to your proof, you wish to prove that
$$forall varepsilon >0exists Nin mathbb N forall nin mathbb Nleft(nge Nimplies left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilonright).$$
Take $varepsilon>0$ and let $N$ be any natural number greater than $dfrac{1+ sqrt{1+32varepsilon}}{8varepsilon}$. Take any natural number $n$ greater than $N$.
Your goal is to prove that $$left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon,$$
which is equivalent to proving $$frac{n+2}{4n^2+2n}<epsilon.$$
You know that $N$ satisfies $fleft(Nright)<hleft(Nright)<varepsilon$, i.e., $N$ satisfies the inequality above. It's easy to prove that because $nge N$, then $n$ will satisfy $f(n)<fleft(Nright)$ and consequently it will satisfy the inequality $left|frac{3n^2+2n+1}{2n^2+n}-frac{3}{2}right|<varepsilon$, as is needed.
Of course Dror's $h$ in his answer is much, much simpler, but my remarks are still compatible with his choice for $h$
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
edited Nov 27 '18 at 13:20
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
answered Mar 2 '14 at 20:10
Git Gud
28.7k1049100
28.7k1049100
add a comment |
add a comment |
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thank you, I'm new to these proofs should have noticed that.
– user123652
Mar 2 '14 at 19:03
@AlijahAhmed What you say is false. It's easy to check for $n=1$.
– Git Gud
Mar 2 '14 at 19:04
I believe you are arguing what you want to prove, in a way. Are you sure the $epsilon - N$ methodology is clear to you?
– user76568
Mar 2 '14 at 19:05
Sorry, completely messed up the factorisation (rather foolish error!)- the fraction $frac{n+2}{4n^2+2n}=frac{(n+2)}{2n(2n+1)}$, see @Dror's answer for a comprehensive answer.
– Alijah Ahmed
Mar 2 '14 at 19:27
@AlijahAhmed I wouldn't call my answer comprehensive... Unless you were being sarcastic..?
– user76568
Mar 2 '14 at 19:50