Mixing
Limits for sequences: Prove that $lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$
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Prove that $$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$$
Note, this is a sequence.
I know that a sequence $(c_n)_n=1^infty$ converges to a finite value L if for every $epsilon > 0$, there exists some $N>0$ so that for all $n$ it holds that:
$$n > N implies |c_n - L| < epsilon$$.
I've learnt this in school: $|c_n - L| < epsilon$ is the distance between each sequence term and the epsilon-value. If we can write $n$ in terms of epsilon, then for any epsilon, we can easily 'choose' an N-value, so that the implication above holds.
I'm afraid I haven't gotten further than this... :
$$left|frac2n^2 - 3n + 59 - n^2 - (-2)right| < epsilon iff left|frac23 - 3n9 - n^2right| < epsilon$$
I just can't seem to reduce $n$ in a way that would help me solve this inequality...
I don't know if there is some algebraic trick that is going over my head or if there is some "theoretical trick" which I could use to simplify. Please, a hint would be greatly appreciated.
sequences-and-series limits
edited Jul 25 at 1:13
Lolita
52318
asked Jul 24 at 23:34
user9750060
13710
add a comment |Â
up vote
2
down vote
favorite
Prove that $$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$$
Note, this is a sequence.
I know that a sequence $(c_n)_n=1^infty$ converges to a finite value L if for every $epsilon > 0$, there exists some $N>0$ so that for all $n$ it holds that:
$$n > N implies |c_n - L| < epsilon$$.
I've learnt this in school: $|c_n - L| < epsilon$ is the distance between each sequence term and the epsilon-value. If we can write $n$ in terms of epsilon, then for any epsilon, we can easily 'choose' an N-value, so that the implication above holds.
I'm afraid I haven't gotten further than this... :
$$left|frac2n^2 - 3n + 59 - n^2 - (-2)right| < epsilon iff left|frac23 - 3n9 - n^2right| < epsilon$$
I just can't seem to reduce $n$ in a way that would help me solve this inequality...
I don't know if there is some algebraic trick that is going over my head or if there is some "theoretical trick" which I could use to simplify. Please, a hint would be greatly appreciated.
sequences-and-series limits
edited Jul 25 at 1:13
Lolita
52318
asked Jul 24 at 23:34
user9750060
13710
1
Advice: Don't start a question with TLDR.
– amWhy
Jul 24 at 23:41
1
Here's a general strategy: Let $f(x) = frac2x^2-3x+59 - x^2$, so that $f(n) = c_n$ when $n = 1 , 2, 3, dots$. Note that $lim_x to infty f(x) = -2$, by say, L'Hopital's Rule, or by dividing top and bottom by $x^2$ (just like in Doug's answer below). If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$ too. Note that this implication only goes one way. If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$, but the converse is false in general.
– JavaMan
Jul 25 at 3:01
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove that $$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$$
Note, this is a sequence.
I know that a sequence $(c_n)_n=1^infty$ converges to a finite value L if for every $epsilon > 0$, there exists some $N>0$ so that for all $n$ it holds that:
$$n > N implies |c_n - L| < epsilon$$.
I've learnt this in school: $|c_n - L| < epsilon$ is the distance between each sequence term and the epsilon-value. If we can write $n$ in terms of epsilon, then for any epsilon, we can easily 'choose' an N-value, so that the implication above holds.
I'm afraid I haven't gotten further than this... :
$$left|frac2n^2 - 3n + 59 - n^2 - (-2)right| < epsilon iff left|frac23 - 3n9 - n^2right| < epsilon$$
I just can't seem to reduce $n$ in a way that would help me solve this inequality...
I don't know if there is some algebraic trick that is going over my head or if there is some "theoretical trick" which I could use to simplify. Please, a hint would be greatly appreciated.
sequences-and-series limits
edited Jul 25 at 1:13
Lolita
52318
asked Jul 24 at 23:34
user9750060
13710
Prove that $$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$$
Note, this is a sequence.
I know that a sequence $(c_n)_n=1^infty$ converges to a finite value L if for every $epsilon > 0$, there exists some $N>0$ so that for all $n$ it holds that:
$$n > N implies |c_n - L| < epsilon$$.
I've learnt this in school: $|c_n - L| < epsilon$ is the distance between each sequence term and the epsilon-value. If we can write $n$ in terms of epsilon, then for any epsilon, we can easily 'choose' an N-value, so that the implication above holds.
I'm afraid I haven't gotten further than this... :
$$left|frac2n^2 - 3n + 59 - n^2 - (-2)right| < epsilon iff left|frac23 - 3n9 - n^2right| < epsilon$$
I just can't seem to reduce $n$ in a way that would help me solve this inequality...
I don't know if there is some algebraic trick that is going over my head or if there is some "theoretical trick" which I could use to simplify. Please, a hint would be greatly appreciated.
sequences-and-series limits
edited Jul 25 at 1:13
Lolita
52318
asked Jul 24 at 23:34
user9750060
13710
edited Jul 25 at 1:13
Lolita
52318
edited Jul 25 at 1:13
Lolita
52318
edited Jul 25 at 1:13
Lolita
52318
52318
asked Jul 24 at 23:34
user9750060
13710
asked Jul 24 at 23:34
user9750060
13710
asked Jul 24 at 23:34
user9750060
13710
13710
1
Advice: Don't start a question with TLDR.
– amWhy
Jul 24 at 23:41
1
Here's a general strategy: Let $f(x) = frac2x^2-3x+59 - x^2$, so that $f(n) = c_n$ when $n = 1 , 2, 3, dots$. Note that $lim_x to infty f(x) = -2$, by say, L'Hopital's Rule, or by dividing top and bottom by $x^2$ (just like in Doug's answer below). If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$ too. Note that this implication only goes one way. If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$, but the converse is false in general.
– JavaMan
Jul 25 at 3:01
add a comment |Â
1
Advice: Don't start a question with TLDR.
– amWhy
Jul 24 at 23:41
1
Here's a general strategy: Let $f(x) = frac2x^2-3x+59 - x^2$, so that $f(n) = c_n$ when $n = 1 , 2, 3, dots$. Note that $lim_x to infty f(x) = -2$, by say, L'Hopital's Rule, or by dividing top and bottom by $x^2$ (just like in Doug's answer below). If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$ too. Note that this implication only goes one way. If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$, but the converse is false in general.
– JavaMan
Jul 25 at 3:01
1
1
Advice: Don't start a question with TLDR.
– amWhy
Jul 24 at 23:41
Advice: Don't start a question with TLDR.
– amWhy
Jul 24 at 23:41
1
1
Here's a general strategy: Let $f(x) = frac2x^2-3x+59 - x^2$, so that $f(n) = c_n$ when $n = 1 , 2, 3, dots$. Note that $lim_x to infty f(x) = -2$, by say, L'Hopital's Rule, or by dividing top and bottom by $x^2$ (just like in Doug's answer below). If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$ too. Note that this implication only goes one way. If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$, but the converse is false in general.
– JavaMan
Jul 25 at 3:01
Here's a general strategy: Let $f(x) = frac2x^2-3x+59 - x^2$, so that $f(n) = c_n$ when $n = 1 , 2, 3, dots$. Note that $lim_x to infty f(x) = -2$, by say, L'Hopital's Rule, or by dividing top and bottom by $x^2$ (just like in Doug's answer below). If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$ too. Note that this implication only goes one way. If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$, but the converse is false in general.
– JavaMan
Jul 25 at 3:01
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
The quick way:
$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$
Divide to and bottom by $n^2$
$lim_n rightarrow infty frac2 - frac 3n + frac 5n^2- 1 + frac 9n^2 = -2$
And when $n$ gets to be large the $frac 1n$ terms become small. The $epsilon-delta$ definition is not necessry.
But if that is the direction you want to go.
There exists an $N$ such that $n>N$ implies $left|frac 3n-23n^2-9right| < epsilon$
let $N = max (frac 3epsilon,4)$
answered Jul 24 at 23:47
Doug M
39k31749
Great thanks! One more question - when should I use 'the quick way' and when should I use epsilon-delta? I don't exactly understand what 'proving a limit' means...
– user9750060
Jul 24 at 23:50
add a comment |Â
up vote
1
down vote
Hint: prove that $frac23-3n9-n^2to 0$.
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
I understand that this proves that the sequence converges, but how will the condition n > N be met? I was taught that we have to write n in terms of epsilon so that for any given epsilon, it would be simple to determine N.
– user9750060
Jul 24 at 23:43
Should I use the comparison theorem to prove that the expression converges and then use my comparison to get epsilon in terms of n?
– user9750060
Jul 24 at 23:47
add a comment |Â
up vote
1
down vote
You want so show that $$left|frac23 - 3n9 - n^2right| < epsilon$$ First when $n$ is large enough ($n>8$) $$left|frac23 - 3n9 - n^2right| =frac3n-23n^2-9<frac3nn^2-9
$$
Also, if $n$ is this large, we have $n^2-9>left(fracn2right)^2$ so $$frac3nn^2-9<frac12nn^2=frac12n$$
Therefore, it's enough to take $$n > maxleft(8,frac12epsilonright)$$
NOTE
I don't recommend doing it this way. You should just show the fraction goes to $0$ which guarantees the existence of such an $N$. Doing it this way is more tedious, no more rigorous, and in my view, adds nothing to understanding.
answered Jul 24 at 23:54
saulspatz
10.4k21323
add a comment |Â
up vote
1
down vote
If we take $n > 3+frac6varepsilon$ we have $ frac3n-3 < fracvarepsilon2$, and if we take $n > sqrtfrac28varepsilon + 9$ we have $ frac14n^2-9 < fracvarepsilon2$.
Therefore for $n ge maxleft4, 3+frac6varepsilon, sqrtfrac28varepsilon + 9right$ we have
beginalign
left|frac2n^2-3n+59-n^2+2right| &= frac23-3n\
&le frac3n+23(n+3)(n-3)\
&= frac3(n+3)+14(n+3)(n-3)\
&= frac3n-3 + frac14n^2-9\
&< fracvarepsilon2 + fracvarepsilon2\
&= varepsilon
endalign
answered Jul 24 at 23:54
mechanodroid
22.2k52041
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The quick way:
$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$
Divide to and bottom by $n^2$
$lim_n rightarrow infty frac2 - frac 3n + frac 5n^2- 1 + frac 9n^2 = -2$
And when $n$ gets to be large the $frac 1n$ terms become small. The $epsilon-delta$ definition is not necessry.
But if that is the direction you want to go.
There exists an $N$ such that $n>N$ implies $left|frac 3n-23n^2-9right| < epsilon$
let $N = max (frac 3epsilon,4)$
answered Jul 24 at 23:47
Doug M
39k31749
Great thanks! One more question - when should I use 'the quick way' and when should I use epsilon-delta? I don't exactly understand what 'proving a limit' means...
– user9750060
Jul 24 at 23:50
add a comment |Â
up vote
1
down vote
accepted
The quick way:
$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$
Divide to and bottom by $n^2$
$lim_n rightarrow infty frac2 - frac 3n + frac 5n^2- 1 + frac 9n^2 = -2$
And when $n$ gets to be large the $frac 1n$ terms become small. The $epsilon-delta$ definition is not necessry.
But if that is the direction you want to go.
There exists an $N$ such that $n>N$ implies $left|frac 3n-23n^2-9right| < epsilon$
let $N = max (frac 3epsilon,4)$
answered Jul 24 at 23:47
Doug M
39k31749
Great thanks! One more question - when should I use 'the quick way' and when should I use epsilon-delta? I don't exactly understand what 'proving a limit' means...
– user9750060
Jul 24 at 23:50
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The quick way:
$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$
Divide to and bottom by $n^2$
$lim_n rightarrow infty frac2 - frac 3n + frac 5n^2- 1 + frac 9n^2 = -2$
And when $n$ gets to be large the $frac 1n$ terms become small. The $epsilon-delta$ definition is not necessry.
But if that is the direction you want to go.
There exists an $N$ such that $n>N$ implies $left|frac 3n-23n^2-9right| < epsilon$
let $N = max (frac 3epsilon,4)$
answered Jul 24 at 23:47
Doug M
39k31749
The quick way:
$lim_n rightarrow infty frac2n^2 - 3n + 59 - n^2 = -2$
Divide to and bottom by $n^2$
$lim_n rightarrow infty frac2 - frac 3n + frac 5n^2- 1 + frac 9n^2 = -2$
And when $n$ gets to be large the $frac 1n$ terms become small. The $epsilon-delta$ definition is not necessry.
But if that is the direction you want to go.
There exists an $N$ such that $n>N$ implies $left|frac 3n-23n^2-9right| < epsilon$
let $N = max (frac 3epsilon,4)$
answered Jul 24 at 23:47
Doug M
39k31749
answered Jul 24 at 23:47
Doug M
39k31749
answered Jul 24 at 23:47
Doug M
39k31749
answered Jul 24 at 23:47
Doug M
39k31749
39k31749
Great thanks! One more question - when should I use 'the quick way' and when should I use epsilon-delta? I don't exactly understand what 'proving a limit' means...
– user9750060
Jul 24 at 23:50
add a comment |Â
Great thanks! One more question - when should I use 'the quick way' and when should I use epsilon-delta? I don't exactly understand what 'proving a limit' means...
– user9750060
Jul 24 at 23:50
Great thanks! One more question - when should I use 'the quick way' and when should I use epsilon-delta? I don't exactly understand what 'proving a limit' means...
– user9750060
Jul 24 at 23:50
Great thanks! One more question - when should I use 'the quick way' and when should I use epsilon-delta? I don't exactly understand what 'proving a limit' means...
– user9750060
Jul 24 at 23:50
add a comment |Â
up vote
1
down vote
Hint: prove that $frac23-3n9-n^2to 0$.
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
I understand that this proves that the sequence converges, but how will the condition n > N be met? I was taught that we have to write n in terms of epsilon so that for any given epsilon, it would be simple to determine N.
– user9750060
Jul 24 at 23:43
Should I use the comparison theorem to prove that the expression converges and then use my comparison to get epsilon in terms of n?
– user9750060
Jul 24 at 23:47
add a comment |Â
up vote
1
down vote
Hint: prove that $frac23-3n9-n^2to 0$.
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
I understand that this proves that the sequence converges, but how will the condition n > N be met? I was taught that we have to write n in terms of epsilon so that for any given epsilon, it would be simple to determine N.
– user9750060
Jul 24 at 23:43
Should I use the comparison theorem to prove that the expression converges and then use my comparison to get epsilon in terms of n?
– user9750060
Jul 24 at 23:47
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: prove that $frac23-3n9-n^2to 0$.
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
Hint: prove that $frac23-3n9-n^2to 0$.
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
answered Jul 24 at 23:38
MartÃn Vacas Vignolo
3,418421
3,418421
I understand that this proves that the sequence converges, but how will the condition n > N be met? I was taught that we have to write n in terms of epsilon so that for any given epsilon, it would be simple to determine N.
– user9750060
Jul 24 at 23:43
Should I use the comparison theorem to prove that the expression converges and then use my comparison to get epsilon in terms of n?
– user9750060
Jul 24 at 23:47
add a comment |Â
I understand that this proves that the sequence converges, but how will the condition n > N be met? I was taught that we have to write n in terms of epsilon so that for any given epsilon, it would be simple to determine N.
– user9750060
Jul 24 at 23:43
Should I use the comparison theorem to prove that the expression converges and then use my comparison to get epsilon in terms of n?
– user9750060
Jul 24 at 23:47
I understand that this proves that the sequence converges, but how will the condition n > N be met? I was taught that we have to write n in terms of epsilon so that for any given epsilon, it would be simple to determine N.
– user9750060
Jul 24 at 23:43
I understand that this proves that the sequence converges, but how will the condition n > N be met? I was taught that we have to write n in terms of epsilon so that for any given epsilon, it would be simple to determine N.
– user9750060
Jul 24 at 23:43
Should I use the comparison theorem to prove that the expression converges and then use my comparison to get epsilon in terms of n?
– user9750060
Jul 24 at 23:47
Should I use the comparison theorem to prove that the expression converges and then use my comparison to get epsilon in terms of n?
– user9750060
Jul 24 at 23:47
add a comment |Â
up vote
1
down vote
You want so show that $$left|frac23 - 3n9 - n^2right| < epsilon$$ First when $n$ is large enough ($n>8$) $$left|frac23 - 3n9 - n^2right| =frac3n-23n^2-9<frac3nn^2-9
$$
Also, if $n$ is this large, we have $n^2-9>left(fracn2right)^2$ so $$frac3nn^2-9<frac12nn^2=frac12n$$
Therefore, it's enough to take $$n > maxleft(8,frac12epsilonright)$$
NOTE
I don't recommend doing it this way. You should just show the fraction goes to $0$ which guarantees the existence of such an $N$. Doing it this way is more tedious, no more rigorous, and in my view, adds nothing to understanding.
answered Jul 24 at 23:54
saulspatz
10.4k21323
add a comment |Â
up vote
1
down vote
You want so show that $$left|frac23 - 3n9 - n^2right| < epsilon$$ First when $n$ is large enough ($n>8$) $$left|frac23 - 3n9 - n^2right| =frac3n-23n^2-9<frac3nn^2-9
$$
Also, if $n$ is this large, we have $n^2-9>left(fracn2right)^2$ so $$frac3nn^2-9<frac12nn^2=frac12n$$
Therefore, it's enough to take $$n > maxleft(8,frac12epsilonright)$$
NOTE
I don't recommend doing it this way. You should just show the fraction goes to $0$ which guarantees the existence of such an $N$. Doing it this way is more tedious, no more rigorous, and in my view, adds nothing to understanding.
answered Jul 24 at 23:54
saulspatz
10.4k21323
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You want so show that $$left|frac23 - 3n9 - n^2right| < epsilon$$ First when $n$ is large enough ($n>8$) $$left|frac23 - 3n9 - n^2right| =frac3n-23n^2-9<frac3nn^2-9
$$
Also, if $n$ is this large, we have $n^2-9>left(fracn2right)^2$ so $$frac3nn^2-9<frac12nn^2=frac12n$$
Therefore, it's enough to take $$n > maxleft(8,frac12epsilonright)$$
NOTE
I don't recommend doing it this way. You should just show the fraction goes to $0$ which guarantees the existence of such an $N$. Doing it this way is more tedious, no more rigorous, and in my view, adds nothing to understanding.
answered Jul 24 at 23:54
saulspatz
10.4k21323
You want so show that $$left|frac23 - 3n9 - n^2right| < epsilon$$ First when $n$ is large enough ($n>8$) $$left|frac23 - 3n9 - n^2right| =frac3n-23n^2-9<frac3nn^2-9
$$
Also, if $n$ is this large, we have $n^2-9>left(fracn2right)^2$ so $$frac3nn^2-9<frac12nn^2=frac12n$$
Therefore, it's enough to take $$n > maxleft(8,frac12epsilonright)$$
NOTE
I don't recommend doing it this way. You should just show the fraction goes to $0$ which guarantees the existence of such an $N$. Doing it this way is more tedious, no more rigorous, and in my view, adds nothing to understanding.
answered Jul 24 at 23:54
saulspatz
10.4k21323
answered Jul 24 at 23:54
saulspatz
10.4k21323
answered Jul 24 at 23:54
saulspatz
10.4k21323
answered Jul 24 at 23:54
saulspatz
10.4k21323
10.4k21323
add a comment |Â
add a comment |Â
up vote
1
down vote
If we take $n > 3+frac6varepsilon$ we have $ frac3n-3 < fracvarepsilon2$, and if we take $n > sqrtfrac28varepsilon + 9$ we have $ frac14n^2-9 < fracvarepsilon2$.
Therefore for $n ge maxleft4, 3+frac6varepsilon, sqrtfrac28varepsilon + 9right$ we have
beginalign
left|frac2n^2-3n+59-n^2+2right| &= frac23-3n\
&le frac3n+23(n+3)(n-3)\
&= frac3(n+3)+14(n+3)(n-3)\
&= frac3n-3 + frac14n^2-9\
&< fracvarepsilon2 + fracvarepsilon2\
&= varepsilon
endalign
answered Jul 24 at 23:54
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
If we take $n > 3+frac6varepsilon$ we have $ frac3n-3 < fracvarepsilon2$, and if we take $n > sqrtfrac28varepsilon + 9$ we have $ frac14n^2-9 < fracvarepsilon2$.
Therefore for $n ge maxleft4, 3+frac6varepsilon, sqrtfrac28varepsilon + 9right$ we have
beginalign
left|frac2n^2-3n+59-n^2+2right| &= frac23-3n\
&le frac3n+23(n+3)(n-3)\
&= frac3(n+3)+14(n+3)(n-3)\
&= frac3n-3 + frac14n^2-9\
&< fracvarepsilon2 + fracvarepsilon2\
&= varepsilon
endalign
answered Jul 24 at 23:54
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If we take $n > 3+frac6varepsilon$ we have $ frac3n-3 < fracvarepsilon2$, and if we take $n > sqrtfrac28varepsilon + 9$ we have $ frac14n^2-9 < fracvarepsilon2$.
Therefore for $n ge maxleft4, 3+frac6varepsilon, sqrtfrac28varepsilon + 9right$ we have
beginalign
left|frac2n^2-3n+59-n^2+2right| &= frac23-3n\
&le frac3n+23(n+3)(n-3)\
&= frac3(n+3)+14(n+3)(n-3)\
&= frac3n-3 + frac14n^2-9\
&< fracvarepsilon2 + fracvarepsilon2\
&= varepsilon
endalign
answered Jul 24 at 23:54
mechanodroid
22.2k52041
If we take $n > 3+frac6varepsilon$ we have $ frac3n-3 < fracvarepsilon2$, and if we take $n > sqrtfrac28varepsilon + 9$ we have $ frac14n^2-9 < fracvarepsilon2$.
Therefore for $n ge maxleft4, 3+frac6varepsilon, sqrtfrac28varepsilon + 9right$ we have
beginalign
left|frac2n^2-3n+59-n^2+2right| &= frac23-3n\
&le frac3n+23(n+3)(n-3)\
&= frac3(n+3)+14(n+3)(n-3)\
&= frac3n-3 + frac14n^2-9\
&< fracvarepsilon2 + fracvarepsilon2\
&= varepsilon
endalign
answered Jul 24 at 23:54
mechanodroid
22.2k52041
answered Jul 24 at 23:54
mechanodroid
22.2k52041
answered Jul 24 at 23:54
mechanodroid
22.2k52041
answered Jul 24 at 23:54
mechanodroid
22.2k52041
22.2k52041
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1
Advice: Don't start a question with TLDR.
– amWhy
Jul 24 at 23:41
1
Here's a general strategy: Let $f(x) = frac2x^2-3x+59 - x^2$, so that $f(n) = c_n$ when $n = 1 , 2, 3, dots$. Note that $lim_x to infty f(x) = -2$, by say, L'Hopital's Rule, or by dividing top and bottom by $x^2$ (just like in Doug's answer below). If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$ too. Note that this implication only goes one way. If $lim_x to infty f(x) = L$, then $lim_n to infty c_n = L$, but the converse is false in general.
– JavaMan
Jul 25 at 3:01