$lim limits_{n to infty }n(sqrt[n]n-1)$ [closed]
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$lim limits_{n to infty }n(sqrt[n]n-1)$
How should I start?
limits limits-without-lhopital
asked Nov 18 at 11:20
matematiccc
1125
closed as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh Nov 20 at 1:14
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-3
down vote
favorite
1
$lim limits_{n to infty }n(sqrt[n]n-1)$
How should I start?
limits limits-without-lhopital
asked Nov 18 at 11:20
matematiccc
1125
closed as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh Nov 20 at 1:14
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
It would be useful if you can show your work here, I can take a look to that.
– gimusi
Nov 18 at 11:43
Try proving that $sqrt[n]{n}-1$ is approximately $frac{log n}{n}$, for instance through the AM-GM inequality and the convexity of $exp$.
– Jack D'Aurizio
Nov 18 at 16:31
add a comment |
up vote
-3
down vote
favorite
1
up vote
-3
down vote
favorite
1
1
$lim limits_{n to infty }n(sqrt[n]n-1)$
How should I start?
limits limits-without-lhopital
asked Nov 18 at 11:20
matematiccc
1125
$lim limits_{n to infty }n(sqrt[n]n-1)$
How should I start?
limits limits-without-lhopital
limits limits-without-lhopital
asked Nov 18 at 11:20
matematiccc
1125
asked Nov 18 at 11:20
matematiccc
1125
asked Nov 18 at 11:20
matematiccc
1125
asked Nov 18 at 11:20
matematiccc
1125
asked Nov 18 at 11:20
matematiccc
1125
1125
closed as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh Nov 20 at 1:14
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh Nov 20 at 1:14
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, amWhy, Rebellos, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
It would be useful if you can show your work here, I can take a look to that.
– gimusi
Nov 18 at 11:43
Try proving that $sqrt[n]{n}-1$ is approximately $frac{log n}{n}$, for instance through the AM-GM inequality and the convexity of $exp$.
– Jack D'Aurizio
Nov 18 at 16:31
add a comment |
It would be useful if you can show your work here, I can take a look to that.
– gimusi
Nov 18 at 11:43
Try proving that $sqrt[n]{n}-1$ is approximately $frac{log n}{n}$, for instance through the AM-GM inequality and the convexity of $exp$.
– Jack D'Aurizio
Nov 18 at 16:31
It would be useful if you can show your work here, I can take a look to that.
– gimusi
Nov 18 at 11:43
It would be useful if you can show your work here, I can take a look to that.
– gimusi
Nov 18 at 11:43
Try proving that $sqrt[n]{n}-1$ is approximately $frac{log n}{n}$, for instance through the AM-GM inequality and the convexity of $exp$.
– Jack D'Aurizio
Nov 18 at 16:31
Try proving that $sqrt[n]{n}-1$ is approximately $frac{log n}{n}$, for instance through the AM-GM inequality and the convexity of $exp$.
– Jack D'Aurizio
Nov 18 at 16:31
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Use that
- $large sqrt[n]n=e^frac{log n}{n}$
- $frac{log n}{n}to 0$
- $t to 0 quad frac{e^t-1}t to 1$
that is
$$n(sqrt[n]n-1)=nleft(e^{frac{log n}n}-1right)=$$$$=log nfrac{e^{frac{log n}n}-1}{frac{log n}n} to infty$$
answered Nov 18 at 11:28
gimusi
89.4k74495
2
Please do read what I wrote. And I proposed L'H only to confirm the actual limit is infinity...
– DonAntonio
Nov 18 at 12:48
1
I wasn't double checking. In fact I was trying to tackle this problem without L'H and also without Taylor series, while I already knew the limit is $;infty;$, and then your answer popped up.
– DonAntonio
Nov 18 at 12:58
@DonAntonio I’ve added a step more to make the hint more clear.
– gimusi
Nov 18 at 15:43
@DonAntonio When I claim that you are missing something I was referring to the fact that you are claiming in the first comment that I was hinting towards a wrong evaluation of the limit. I hope that now my aim is clear.
– gimusi
Nov 18 at 18:50
That's clear now . Had you said that in your first comment, which speaks of L'H when I didn't speak of it, It would have been clear then. I am deleting my comment above now and +1
– DonAntonio
Nov 19 at 0:05
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Use that
- $large sqrt[n]n=e^frac{log n}{n}$
- $frac{log n}{n}to 0$
- $t to 0 quad frac{e^t-1}t to 1$
that is
$$n(sqrt[n]n-1)=nleft(e^{frac{log n}n}-1right)=$$$$=log nfrac{e^{frac{log n}n}-1}{frac{log n}n} to infty$$
answered Nov 18 at 11:28
gimusi
89.4k74495
2
Please do read what I wrote. And I proposed L'H only to confirm the actual limit is infinity...
– DonAntonio
Nov 18 at 12:48
1
I wasn't double checking. In fact I was trying to tackle this problem without L'H and also without Taylor series, while I already knew the limit is $;infty;$, and then your answer popped up.
– DonAntonio
Nov 18 at 12:58
@DonAntonio I’ve added a step more to make the hint more clear.
– gimusi
Nov 18 at 15:43
@DonAntonio When I claim that you are missing something I was referring to the fact that you are claiming in the first comment that I was hinting towards a wrong evaluation of the limit. I hope that now my aim is clear.
– gimusi
Nov 18 at 18:50
That's clear now . Had you said that in your first comment, which speaks of L'H when I didn't speak of it, It would have been clear then. I am deleting my comment above now and +1
– DonAntonio
Nov 19 at 0:05
|
show 1 more comment
up vote
2
down vote
accepted
Use that
- $large sqrt[n]n=e^frac{log n}{n}$
- $frac{log n}{n}to 0$
- $t to 0 quad frac{e^t-1}t to 1$
that is
$$n(sqrt[n]n-1)=nleft(e^{frac{log n}n}-1right)=$$$$=log nfrac{e^{frac{log n}n}-1}{frac{log n}n} to infty$$
answered Nov 18 at 11:28
gimusi
89.4k74495
2
Please do read what I wrote. And I proposed L'H only to confirm the actual limit is infinity...
– DonAntonio
Nov 18 at 12:48
1
I wasn't double checking. In fact I was trying to tackle this problem without L'H and also without Taylor series, while I already knew the limit is $;infty;$, and then your answer popped up.
– DonAntonio
Nov 18 at 12:58
@DonAntonio I’ve added a step more to make the hint more clear.
– gimusi
Nov 18 at 15:43
@DonAntonio When I claim that you are missing something I was referring to the fact that you are claiming in the first comment that I was hinting towards a wrong evaluation of the limit. I hope that now my aim is clear.
– gimusi
Nov 18 at 18:50
That's clear now . Had you said that in your first comment, which speaks of L'H when I didn't speak of it, It would have been clear then. I am deleting my comment above now and +1
– DonAntonio
Nov 19 at 0:05
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Use that
- $large sqrt[n]n=e^frac{log n}{n}$
- $frac{log n}{n}to 0$
- $t to 0 quad frac{e^t-1}t to 1$
that is
$$n(sqrt[n]n-1)=nleft(e^{frac{log n}n}-1right)=$$$$=log nfrac{e^{frac{log n}n}-1}{frac{log n}n} to infty$$
answered Nov 18 at 11:28
gimusi
89.4k74495
Use that
- $large sqrt[n]n=e^frac{log n}{n}$
- $frac{log n}{n}to 0$
- $t to 0 quad frac{e^t-1}t to 1$
that is
$$n(sqrt[n]n-1)=nleft(e^{frac{log n}n}-1right)=$$$$=log nfrac{e^{frac{log n}n}-1}{frac{log n}n} to infty$$
answered Nov 18 at 11:28
gimusi
89.4k74495
edited Nov 19 at 19:01
answered Nov 18 at 11:28
gimusi
89.4k74495
answered Nov 18 at 11:28
gimusi
89.4k74495
answered Nov 18 at 11:28
gimusi
89.4k74495
89.4k74495
2
Please do read what I wrote. And I proposed L'H only to confirm the actual limit is infinity...
– DonAntonio
Nov 18 at 12:48
1
I wasn't double checking. In fact I was trying to tackle this problem without L'H and also without Taylor series, while I already knew the limit is $;infty;$, and then your answer popped up.
– DonAntonio
Nov 18 at 12:58
@DonAntonio I’ve added a step more to make the hint more clear.
– gimusi
Nov 18 at 15:43
@DonAntonio When I claim that you are missing something I was referring to the fact that you are claiming in the first comment that I was hinting towards a wrong evaluation of the limit. I hope that now my aim is clear.
– gimusi
Nov 18 at 18:50
That's clear now . Had you said that in your first comment, which speaks of L'H when I didn't speak of it, It would have been clear then. I am deleting my comment above now and +1
– DonAntonio
Nov 19 at 0:05
|
show 1 more comment
2
Please do read what I wrote. And I proposed L'H only to confirm the actual limit is infinity...
– DonAntonio
Nov 18 at 12:48
1
I wasn't double checking. In fact I was trying to tackle this problem without L'H and also without Taylor series, while I already knew the limit is $;infty;$, and then your answer popped up.
– DonAntonio
Nov 18 at 12:58
@DonAntonio I’ve added a step more to make the hint more clear.
– gimusi
Nov 18 at 15:43
@DonAntonio When I claim that you are missing something I was referring to the fact that you are claiming in the first comment that I was hinting towards a wrong evaluation of the limit. I hope that now my aim is clear.
– gimusi
Nov 18 at 18:50
That's clear now . Had you said that in your first comment, which speaks of L'H when I didn't speak of it, It would have been clear then. I am deleting my comment above now and +1
– DonAntonio
Nov 19 at 0:05
2
2
Please do read what I wrote. And I proposed L'H only to confirm the actual limit is infinity...
– DonAntonio
Nov 18 at 12:48
Please do read what I wrote. And I proposed L'H only to confirm the actual limit is infinity...
– DonAntonio
Nov 18 at 12:48
1
1
I wasn't double checking. In fact I was trying to tackle this problem without L'H and also without Taylor series, while I already knew the limit is $;infty;$, and then your answer popped up.
– DonAntonio
Nov 18 at 12:58
I wasn't double checking. In fact I was trying to tackle this problem without L'H and also without Taylor series, while I already knew the limit is $;infty;$, and then your answer popped up.
– DonAntonio
Nov 18 at 12:58
@DonAntonio I’ve added a step more to make the hint more clear.
– gimusi
Nov 18 at 15:43
@DonAntonio I’ve added a step more to make the hint more clear.
– gimusi
Nov 18 at 15:43
@DonAntonio When I claim that you are missing something I was referring to the fact that you are claiming in the first comment that I was hinting towards a wrong evaluation of the limit. I hope that now my aim is clear.
– gimusi
Nov 18 at 18:50
@DonAntonio When I claim that you are missing something I was referring to the fact that you are claiming in the first comment that I was hinting towards a wrong evaluation of the limit. I hope that now my aim is clear.
– gimusi
Nov 18 at 18:50
That's clear now . Had you said that in your first comment, which speaks of L'H when I didn't speak of it, It would have been clear then. I am deleting my comment above now and +1
– DonAntonio
Nov 19 at 0:05
That's clear now . Had you said that in your first comment, which speaks of L'H when I didn't speak of it, It would have been clear then. I am deleting my comment above now and +1
– DonAntonio
Nov 19 at 0:05
|
show 1 more comment
It would be useful if you can show your work here, I can take a look to that.
– gimusi
Nov 18 at 11:43
Try proving that $sqrt[n]{n}-1$ is approximately $frac{log n}{n}$, for instance through the AM-GM inequality and the convexity of $exp$.
– Jack D'Aurizio
Nov 18 at 16:31