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How do I evaluate this limit, using Sandwich Theorem? [duplicate]
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This question already has an answer here:
Show that $lim_{nrightarrow infty} sqrt[n]{c_1^n+c_2^n+ldots+c_m^n} = max{c_1,c_2,ldots,c_m}$ [duplicate]
3 answers
Using Sandwich Theorem, how can I evaluate the limit for -
$$lim_{nto infty} (a^n+b^n)^{1/n}$$ where $a$ and $b$ are positive quantities and $b$ is greater than $a$.
limits limits-without-lhopital graph-limits
edited Jan 18 at 6:23
Seewoo Lee
6,959927
asked Jan 18 at 6:03
user574937user574937
384
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marked as duplicate by Hans Lundmark, lab bhattacharjee limits
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Jan 18 at 7:04
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This question already has an answer here:
Show that $lim_{nrightarrow infty} sqrt[n]{c_1^n+c_2^n+ldots+c_m^n} = max{c_1,c_2,ldots,c_m}$ [duplicate]
3 answers
Using Sandwich Theorem, how can I evaluate the limit for -
$$lim_{nto infty} (a^n+b^n)^{1/n}$$ where $a$ and $b$ are positive quantities and $b$ is greater than $a$.
limits limits-without-lhopital graph-limits
edited Jan 18 at 6:23
Seewoo Lee
6,959927
asked Jan 18 at 6:03
user574937user574937
384
$endgroup$
marked as duplicate by Hans Lundmark, lab bhattacharjee limits
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Jan 18 at 7:04
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This question already has an answer here:
Show that $lim_{nrightarrow infty} sqrt[n]{c_1^n+c_2^n+ldots+c_m^n} = max{c_1,c_2,ldots,c_m}$ [duplicate]
3 answers
Using Sandwich Theorem, how can I evaluate the limit for -
$$lim_{nto infty} (a^n+b^n)^{1/n}$$ where $a$ and $b$ are positive quantities and $b$ is greater than $a$.
limits limits-without-lhopital graph-limits
edited Jan 18 at 6:23
Seewoo Lee
6,959927
asked Jan 18 at 6:03
user574937user574937
384
$endgroup$
This question already has an answer here:
Show that $lim_{nrightarrow infty} sqrt[n]{c_1^n+c_2^n+ldots+c_m^n} = max{c_1,c_2,ldots,c_m}$ [duplicate]
3 answers
Using Sandwich Theorem, how can I evaluate the limit for -
$$lim_{nto infty} (a^n+b^n)^{1/n}$$ where $a$ and $b$ are positive quantities and $b$ is greater than $a$.
This question already has an answer here:
Show that $lim_{nrightarrow infty} sqrt[n]{c_1^n+c_2^n+ldots+c_m^n} = max{c_1,c_2,ldots,c_m}$ [duplicate]
3 answers
limits limits-without-lhopital graph-limits
limits limits-without-lhopital graph-limits
edited Jan 18 at 6:23
Seewoo Lee
6,959927
asked Jan 18 at 6:03
user574937user574937
384
edited Jan 18 at 6:23
Seewoo Lee
6,959927
asked Jan 18 at 6:03
user574937user574937
384
edited Jan 18 at 6:23
Seewoo Lee
6,959927
edited Jan 18 at 6:23
Seewoo Lee
6,959927
edited Jan 18 at 6:23
Seewoo Lee
6,959927
6,959927
asked Jan 18 at 6:03
user574937user574937
384
asked Jan 18 at 6:03
user574937user574937
384
asked Jan 18 at 6:03
user574937user574937
384
384
marked as duplicate by Hans Lundmark, lab bhattacharjee limits
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Jan 18 at 7:04
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1 Answer
1
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oldest
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4
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$$b < (a^{n} + b^{n})^{1/n} < 2^{1/n}b$$
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
$begingroup$
Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out.
$endgroup$
– user574937
Jan 18 at 18:36
$begingroup$
@user574937 If you can show that the inequality is true, than we get the result from $lim_{ntoinfty} 2^{1/n} = 2^{lim_{nto infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$.
$endgroup$
– Seewoo Lee
Jan 18 at 22:24
$begingroup$
That part was understood. I mean, how did you get upon this conclusion? What and how did you thought?
$endgroup$
– user574937
Jan 20 at 4:41
$begingroup$
@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit.
$endgroup$
– Seewoo Lee
Jan 20 at 4:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
$$b < (a^{n} + b^{n})^{1/n} < 2^{1/n}b$$
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
$begingroup$
Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out.
$endgroup$
– user574937
Jan 18 at 18:36
$begingroup$
@user574937 If you can show that the inequality is true, than we get the result from $lim_{ntoinfty} 2^{1/n} = 2^{lim_{nto infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$.
$endgroup$
– Seewoo Lee
Jan 18 at 22:24
$begingroup$
That part was understood. I mean, how did you get upon this conclusion? What and how did you thought?
$endgroup$
– user574937
Jan 20 at 4:41
$begingroup$
@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit.
$endgroup$
– Seewoo Lee
Jan 20 at 4:59
add a comment |
4
$begingroup$
$$b < (a^{n} + b^{n})^{1/n} < 2^{1/n}b$$
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
$begingroup$
Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out.
$endgroup$
– user574937
Jan 18 at 18:36
$begingroup$
@user574937 If you can show that the inequality is true, than we get the result from $lim_{ntoinfty} 2^{1/n} = 2^{lim_{nto infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$.
$endgroup$
– Seewoo Lee
Jan 18 at 22:24
$begingroup$
That part was understood. I mean, how did you get upon this conclusion? What and how did you thought?
$endgroup$
– user574937
Jan 20 at 4:41
$begingroup$
@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit.
$endgroup$
– Seewoo Lee
Jan 20 at 4:59
add a comment |
4
4
4
$begingroup$
$$b < (a^{n} + b^{n})^{1/n} < 2^{1/n}b$$
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
$$b < (a^{n} + b^{n})^{1/n} < 2^{1/n}b$$
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
answered Jan 18 at 6:24
Seewoo LeeSeewoo Lee
6,959927
6,959927
$begingroup$
Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out.
$endgroup$
– user574937
Jan 18 at 18:36
$begingroup$
@user574937 If you can show that the inequality is true, than we get the result from $lim_{ntoinfty} 2^{1/n} = 2^{lim_{nto infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$.
$endgroup$
– Seewoo Lee
Jan 18 at 22:24
$begingroup$
That part was understood. I mean, how did you get upon this conclusion? What and how did you thought?
$endgroup$
– user574937
Jan 20 at 4:41
$begingroup$
@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit.
$endgroup$
– Seewoo Lee
Jan 20 at 4:59
add a comment |
$begingroup$
Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out.
$endgroup$
– user574937
Jan 18 at 18:36
$begingroup$
@user574937 If you can show that the inequality is true, than we get the result from $lim_{ntoinfty} 2^{1/n} = 2^{lim_{nto infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$.
$endgroup$
– Seewoo Lee
Jan 18 at 22:24
$begingroup$
That part was understood. I mean, how did you get upon this conclusion? What and how did you thought?
$endgroup$
– user574937
Jan 20 at 4:41
$begingroup$
@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit.
$endgroup$
– Seewoo Lee
Jan 20 at 4:59
$begingroup$
Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out.
$endgroup$
– user574937
Jan 18 at 18:36
$begingroup$
Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out.
$endgroup$
– user574937
Jan 18 at 18:36
$begingroup$
@user574937 If you can show that the inequality is true, than we get the result from $lim_{ntoinfty} 2^{1/n} = 2^{lim_{nto infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$.
$endgroup$
– Seewoo Lee
Jan 18 at 22:24
$begingroup$
@user574937 If you can show that the inequality is true, than we get the result from $lim_{ntoinfty} 2^{1/n} = 2^{lim_{nto infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$.
$endgroup$
– Seewoo Lee
Jan 18 at 22:24
$begingroup$
That part was understood. I mean, how did you get upon this conclusion? What and how did you thought?
$endgroup$
– user574937
Jan 20 at 4:41
$begingroup$
That part was understood. I mean, how did you get upon this conclusion? What and how did you thought?
$endgroup$
– user574937
Jan 20 at 4:41
$begingroup$
@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit.
$endgroup$
– Seewoo Lee
Jan 20 at 4:59
$begingroup$
@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit.
$endgroup$
– Seewoo Lee
Jan 20 at 4:59
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