What is the result to $lim_{ntoinfty} sqrt[5]frac{-n^5-2}{n^4+n^2+1} $? [closed]
$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$
I got $-infty$ but my answer sheet says $+infty$ and I don't know if it's just an error in the answers or am I doing something wrong?
limits
edited Nov 25 at 15:43
Henning Makholm
237k16301536
asked Nov 25 at 15:25
agromek
345
closed as off-topic by amWhy, Did, user21820, Jyrki Lahtonen, RRL Dec 2 at 2:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$
I got $-infty$ but my answer sheet says $+infty$ and I don't know if it's just an error in the answers or am I doing something wrong?
limits
edited Nov 25 at 15:43
Henning Makholm
237k16301536
asked Nov 25 at 15:25
agromek
345
closed as off-topic by amWhy, Did, user21820, Jyrki Lahtonen, RRL Dec 2 at 2:17
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." – amWhy, Did, user21820, Jyrki Lahtonen, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
The argument of the root is negative.
– Yves Daoust
Nov 25 at 15:55
You can check you're right here: symbolab.com/solver/limit-calculator/…
– Dr. Mathva
Nov 25 at 16:01
add a comment |
$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$
I got $-infty$ but my answer sheet says $+infty$ and I don't know if it's just an error in the answers or am I doing something wrong?
limits
edited Nov 25 at 15:43
Henning Makholm
237k16301536
asked Nov 25 at 15:25
agromek
345
$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$
I got $-infty$ but my answer sheet says $+infty$ and I don't know if it's just an error in the answers or am I doing something wrong?
limits
limits
edited Nov 25 at 15:43
Henning Makholm
237k16301536
asked Nov 25 at 15:25
agromek
345
edited Nov 25 at 15:43
Henning Makholm
237k16301536
asked Nov 25 at 15:25
agromek
345
edited Nov 25 at 15:43
Henning Makholm
237k16301536
edited Nov 25 at 15:43
Henning Makholm
237k16301536
edited Nov 25 at 15:43
Henning Makholm
237k16301536
237k16301536
asked Nov 25 at 15:25
agromek
345
asked Nov 25 at 15:25
agromek
345
asked Nov 25 at 15:25
agromek
345
345
closed as off-topic by amWhy, Did, user21820, Jyrki Lahtonen, RRL Dec 2 at 2:17
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." – amWhy, Did, user21820, Jyrki Lahtonen, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Did, user21820, Jyrki Lahtonen, RRL Dec 2 at 2:17
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." – amWhy, Did, user21820, Jyrki Lahtonen, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
The argument of the root is negative.
– Yves Daoust
Nov 25 at 15:55
You can check you're right here: symbolab.com/solver/limit-calculator/…
– Dr. Mathva
Nov 25 at 16:01
add a comment |
The argument of the root is negative.
– Yves Daoust
Nov 25 at 15:55
You can check you're right here: symbolab.com/solver/limit-calculator/…
– Dr. Mathva
Nov 25 at 16:01
The argument of the root is negative.
– Yves Daoust
Nov 25 at 15:55
The argument of the root is negative.
– Yves Daoust
Nov 25 at 15:55
You can check you're right here: symbolab.com/solver/limit-calculator/…
– Dr. Mathva
Nov 25 at 16:01
You can check you're right here: symbolab.com/solver/limit-calculator/…
– Dr. Mathva
Nov 25 at 16:01
add a comment |
1 Answer
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You are right indeed
$$ left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) sim sqrt[5]{-n}to -infty$$
Maybe there is a typo in your book or for the limit at $-infty$ or for the result.
answered Nov 25 at 15:41
gimusi
1
That is what I thought, thank you!
– agromek
Nov 25 at 15:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are right indeed
$$ left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) sim sqrt[5]{-n}to -infty$$
Maybe there is a typo in your book or for the limit at $-infty$ or for the result.
answered Nov 25 at 15:41
gimusi
1
That is what I thought, thank you!
– agromek
Nov 25 at 15:43
add a comment |
You are right indeed
$$ left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) sim sqrt[5]{-n}to -infty$$
Maybe there is a typo in your book or for the limit at $-infty$ or for the result.
answered Nov 25 at 15:41
gimusi
1
That is what I thought, thank you!
– agromek
Nov 25 at 15:43
add a comment |
You are right indeed
$$ left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) sim sqrt[5]{-n}to -infty$$
Maybe there is a typo in your book or for the limit at $-infty$ or for the result.
answered Nov 25 at 15:41
gimusi
1
You are right indeed
$$ left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) sim sqrt[5]{-n}to -infty$$
Maybe there is a typo in your book or for the limit at $-infty$ or for the result.
answered Nov 25 at 15:41
gimusi
1
answered Nov 25 at 15:41
gimusi
1
answered Nov 25 at 15:41
gimusi
1
answered Nov 25 at 15:41
gimusi
1
1
That is what I thought, thank you!
– agromek
Nov 25 at 15:43
add a comment |
That is what I thought, thank you!
– agromek
Nov 25 at 15:43
That is what I thought, thank you!
– agromek
Nov 25 at 15:43
That is what I thought, thank you!
– agromek
Nov 25 at 15:43
add a comment |
The argument of the root is negative.
– Yves Daoust
Nov 25 at 15:55
You can check you're right here: symbolab.com/solver/limit-calculator/…
– Dr. Mathva
Nov 25 at 16:01