the limit of two funtions going to infinity converges to a positive constant [closed]
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as following a proof from a book it state as(please see pic if still confusing):
since lim ->∞ (f(n)/g(n)) = c >0, it follows from the definition of a limit that there is some n0 beyond which the ration is always between 1/2 c and 2c. Thus, f(n) <2cg(n) for all n >= n0, which implies that f(n) = O(g(n));
Can someone please explain a bit why there is some n0 beyond which the ratio is always between 1/2c and 2c ? enter image description here
limits proof-explanation
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
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closed as off-topic by Don Thousand, Namaste, egreg, Leucippus, Shailesh Dec 27 '18 at 0:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Don Thousand, Namaste, egreg, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
as following a proof from a book it state as(please see pic if still confusing):
since lim ->∞ (f(n)/g(n)) = c >0, it follows from the definition of a limit that there is some n0 beyond which the ration is always between 1/2 c and 2c. Thus, f(n) <2cg(n) for all n >= n0, which implies that f(n) = O(g(n));
Can someone please explain a bit why there is some n0 beyond which the ratio is always between 1/2c and 2c ? enter image description here
limits proof-explanation
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
$endgroup$
closed as off-topic by Don Thousand, Namaste, egreg, Leucippus, Shailesh Dec 27 '18 at 0:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Don Thousand, Namaste, egreg, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
as following a proof from a book it state as(please see pic if still confusing):
since lim ->∞ (f(n)/g(n)) = c >0, it follows from the definition of a limit that there is some n0 beyond which the ration is always between 1/2 c and 2c. Thus, f(n) <2cg(n) for all n >= n0, which implies that f(n) = O(g(n));
Can someone please explain a bit why there is some n0 beyond which the ratio is always between 1/2c and 2c ? enter image description here
limits proof-explanation
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
$endgroup$
as following a proof from a book it state as(please see pic if still confusing):
since lim ->∞ (f(n)/g(n)) = c >0, it follows from the definition of a limit that there is some n0 beyond which the ration is always between 1/2 c and 2c. Thus, f(n) <2cg(n) for all n >= n0, which implies that f(n) = O(g(n));
Can someone please explain a bit why there is some n0 beyond which the ratio is always between 1/2c and 2c ? enter image description here
limits proof-explanation
limits proof-explanation
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
asked Dec 26 '18 at 18:43
MaxfieldMaxfield
424
424
closed as off-topic by Don Thousand, Namaste, egreg, Leucippus, Shailesh Dec 27 '18 at 0:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Don Thousand, Namaste, egreg, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Don Thousand, Namaste, egreg, Leucippus, Shailesh Dec 27 '18 at 0:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Don Thousand, Namaste, egreg, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
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That's the very definition of $lim_{nto infty} h(n) = c$.
That means for any value $epsilon > 0$ we can find some value $n_0$ so that for all $n > n_0$ ir will always be then case that $c-epsilon < h(n) < c +epsilon$.
There's nothing to explain why, because that is what $lim_{nto infty} h(n) = c$ means.
So $lim_{nto infty} frac {f(n)}{g(n)} = c$ then we can find for ANY $epsilon > 0$, in particular for $epsilon = frac 12 c$, there is some $n_0$ so that for all $n > n_0$ we will always have:
$c - epsilon < frac {f(n)}{g(n)} < c + epsilon$ so
$c - frac 12 c < frac {f(n)}{g(n)} < c + frac 12c $ so
$frac 12 c < frac {f(n)}{g(n)} < frac 32c < 2c$
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
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typo ? it will instead of ir will ?
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– Maxfield
Dec 26 '18 at 20:24
add a comment |
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This is just a quick and dirty version of an $epsilontext{-}delta$ argument; the fact that $c>0$ means that $epsilon_1=c/2>0$ and $epsilon_2=c>0$ are both valid "choices" for the limit definition, and then the rest falls into place.
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's the very definition of $lim_{nto infty} h(n) = c$.
That means for any value $epsilon > 0$ we can find some value $n_0$ so that for all $n > n_0$ ir will always be then case that $c-epsilon < h(n) < c +epsilon$.
There's nothing to explain why, because that is what $lim_{nto infty} h(n) = c$ means.
So $lim_{nto infty} frac {f(n)}{g(n)} = c$ then we can find for ANY $epsilon > 0$, in particular for $epsilon = frac 12 c$, there is some $n_0$ so that for all $n > n_0$ we will always have:
$c - epsilon < frac {f(n)}{g(n)} < c + epsilon$ so
$c - frac 12 c < frac {f(n)}{g(n)} < c + frac 12c $ so
$frac 12 c < frac {f(n)}{g(n)} < frac 32c < 2c$
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
typo ? it will instead of ir will ?
$endgroup$
– Maxfield
Dec 26 '18 at 20:24
add a comment |
$begingroup$
That's the very definition of $lim_{nto infty} h(n) = c$.
That means for any value $epsilon > 0$ we can find some value $n_0$ so that for all $n > n_0$ ir will always be then case that $c-epsilon < h(n) < c +epsilon$.
There's nothing to explain why, because that is what $lim_{nto infty} h(n) = c$ means.
So $lim_{nto infty} frac {f(n)}{g(n)} = c$ then we can find for ANY $epsilon > 0$, in particular for $epsilon = frac 12 c$, there is some $n_0$ so that for all $n > n_0$ we will always have:
$c - epsilon < frac {f(n)}{g(n)} < c + epsilon$ so
$c - frac 12 c < frac {f(n)}{g(n)} < c + frac 12c $ so
$frac 12 c < frac {f(n)}{g(n)} < frac 32c < 2c$
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
typo ? it will instead of ir will ?
$endgroup$
– Maxfield
Dec 26 '18 at 20:24
add a comment |
$begingroup$
That's the very definition of $lim_{nto infty} h(n) = c$.
That means for any value $epsilon > 0$ we can find some value $n_0$ so that for all $n > n_0$ ir will always be then case that $c-epsilon < h(n) < c +epsilon$.
There's nothing to explain why, because that is what $lim_{nto infty} h(n) = c$ means.
So $lim_{nto infty} frac {f(n)}{g(n)} = c$ then we can find for ANY $epsilon > 0$, in particular for $epsilon = frac 12 c$, there is some $n_0$ so that for all $n > n_0$ we will always have:
$c - epsilon < frac {f(n)}{g(n)} < c + epsilon$ so
$c - frac 12 c < frac {f(n)}{g(n)} < c + frac 12c $ so
$frac 12 c < frac {f(n)}{g(n)} < frac 32c < 2c$
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
$endgroup$
That's the very definition of $lim_{nto infty} h(n) = c$.
That means for any value $epsilon > 0$ we can find some value $n_0$ so that for all $n > n_0$ ir will always be then case that $c-epsilon < h(n) < c +epsilon$.
There's nothing to explain why, because that is what $lim_{nto infty} h(n) = c$ means.
So $lim_{nto infty} frac {f(n)}{g(n)} = c$ then we can find for ANY $epsilon > 0$, in particular for $epsilon = frac 12 c$, there is some $n_0$ so that for all $n > n_0$ we will always have:
$c - epsilon < frac {f(n)}{g(n)} < c + epsilon$ so
$c - frac 12 c < frac {f(n)}{g(n)} < c + frac 12c $ so
$frac 12 c < frac {f(n)}{g(n)} < frac 32c < 2c$
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
answered Dec 26 '18 at 19:02
fleabloodfleablood
72.3k22687
72.3k22687
$begingroup$
typo ? it will instead of ir will ?
$endgroup$
– Maxfield
Dec 26 '18 at 20:24
add a comment |
$begingroup$
typo ? it will instead of ir will ?
$endgroup$
– Maxfield
Dec 26 '18 at 20:24
$begingroup$
typo ? it will instead of ir will ?
$endgroup$
– Maxfield
Dec 26 '18 at 20:24
$begingroup$
typo ? it will instead of ir will ?
$endgroup$
– Maxfield
Dec 26 '18 at 20:24
add a comment |
$begingroup$
This is just a quick and dirty version of an $epsilontext{-}delta$ argument; the fact that $c>0$ means that $epsilon_1=c/2>0$ and $epsilon_2=c>0$ are both valid "choices" for the limit definition, and then the rest falls into place.
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
$endgroup$
add a comment |
$begingroup$
This is just a quick and dirty version of an $epsilontext{-}delta$ argument; the fact that $c>0$ means that $epsilon_1=c/2>0$ and $epsilon_2=c>0$ are both valid "choices" for the limit definition, and then the rest falls into place.
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
$endgroup$
add a comment |
$begingroup$
This is just a quick and dirty version of an $epsilontext{-}delta$ argument; the fact that $c>0$ means that $epsilon_1=c/2>0$ and $epsilon_2=c>0$ are both valid "choices" for the limit definition, and then the rest falls into place.
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
$endgroup$
This is just a quick and dirty version of an $epsilontext{-}delta$ argument; the fact that $c>0$ means that $epsilon_1=c/2>0$ and $epsilon_2=c>0$ are both valid "choices" for the limit definition, and then the rest falls into place.
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
answered Dec 26 '18 at 18:47
IAmTheGuyIAmTheGuy
312
312
add a comment |
add a comment |
