What is the result to $lim_{ntoinfty} sqrt[5]frac{-n^5-2}{n^4+n^2+1} $? [closed]












-1














$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$



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







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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

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  • 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
















-1














$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$



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







share|cite|improve this 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














-1












-1








-1







$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$



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







share|cite|improve this question















$lim_{ntoinfty} left( sqrt[5]frac{-n^5-2}{n^4+n^2+1} right) = lim_{ntoinfty} left( sqrt[5]frac{-n-frac{2}{n^4}}{1+frac{1}{n^2}+frac{1}{n^4}} right)=-infty$



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






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share|cite|improve this question













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edited Nov 25 at 15:43









Henning Makholm

237k16301536




237k16301536










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


















  • 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










1 Answer
1






active

oldest

votes


















0














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.






share|cite|improve this answer





















  • That is what I thought, thank you!
    – agromek
    Nov 25 at 15:43


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














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.






share|cite|improve this answer





















  • That is what I thought, thank you!
    – agromek
    Nov 25 at 15:43
















0














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.






share|cite|improve this answer





















  • That is what I thought, thank you!
    – agromek
    Nov 25 at 15:43














0












0








0






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 at 15:41









gimusi

1




1












  • 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




That is what I thought, thank you!
– agromek
Nov 25 at 15:43



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