limit of trigonometric function to infinity












0












$begingroup$


I'm facing a bit of trouble figuring out this limit.



$$ lim_{n to infty} cosleft(left(-1right)^n frac{n-1}{n+1}piright)$$



and I'm not sure if I can simply find the limit of the inner functions and then apply cosine to that, as in
$$ lim_{n to infty} (-1)^n = undefined quad quad lim_{n to infty} frac{n-1}{n+1} = 1 quad quad lim_{n to infty} pi = pi $$
But because of the oscillation caused by $displaystylelim_{n to infty} (-1)^n$, I am not sure what I should do. It would seem to me that the entire thing is undefined, but that is a bad answer.






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    $ntoinfty$ does not necessarily imply $n$ is an integer, for non integer $n,(-1)^n=?$
    $endgroup$
    – lab bhattacharjee
    Apr 28 '18 at 18:11










  • $begingroup$
    True. Then it would tend to oscillate. I'm not quite sure what I can do with that then. Thanks
    $endgroup$
    – user25758
    Apr 28 '18 at 18:20










  • $begingroup$
    Usually $n$ is used to denote an integer, so assuming it is, note that $cos(-pi)=cos(pi)=-1$.
    $endgroup$
    – Cedron Dawg
    Apr 28 '18 at 19:06


















0












$begingroup$


I'm facing a bit of trouble figuring out this limit.



$$ lim_{n to infty} cosleft(left(-1right)^n frac{n-1}{n+1}piright)$$



and I'm not sure if I can simply find the limit of the inner functions and then apply cosine to that, as in
$$ lim_{n to infty} (-1)^n = undefined quad quad lim_{n to infty} frac{n-1}{n+1} = 1 quad quad lim_{n to infty} pi = pi $$
But because of the oscillation caused by $displaystylelim_{n to infty} (-1)^n$, I am not sure what I should do. It would seem to me that the entire thing is undefined, but that is a bad answer.






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    $ntoinfty$ does not necessarily imply $n$ is an integer, for non integer $n,(-1)^n=?$
    $endgroup$
    – lab bhattacharjee
    Apr 28 '18 at 18:11










  • $begingroup$
    True. Then it would tend to oscillate. I'm not quite sure what I can do with that then. Thanks
    $endgroup$
    – user25758
    Apr 28 '18 at 18:20










  • $begingroup$
    Usually $n$ is used to denote an integer, so assuming it is, note that $cos(-pi)=cos(pi)=-1$.
    $endgroup$
    – Cedron Dawg
    Apr 28 '18 at 19:06
















0












0








0


2



$begingroup$


I'm facing a bit of trouble figuring out this limit.



$$ lim_{n to infty} cosleft(left(-1right)^n frac{n-1}{n+1}piright)$$



and I'm not sure if I can simply find the limit of the inner functions and then apply cosine to that, as in
$$ lim_{n to infty} (-1)^n = undefined quad quad lim_{n to infty} frac{n-1}{n+1} = 1 quad quad lim_{n to infty} pi = pi $$
But because of the oscillation caused by $displaystylelim_{n to infty} (-1)^n$, I am not sure what I should do. It would seem to me that the entire thing is undefined, but that is a bad answer.






limits







share|cite|improve this question











$endgroup$




I'm facing a bit of trouble figuring out this limit.



$$ lim_{n to infty} cosleft(left(-1right)^n frac{n-1}{n+1}piright)$$



and I'm not sure if I can simply find the limit of the inner functions and then apply cosine to that, as in
$$ lim_{n to infty} (-1)^n = undefined quad quad lim_{n to infty} frac{n-1}{n+1} = 1 quad quad lim_{n to infty} pi = pi $$
But because of the oscillation caused by $displaystylelim_{n to infty} (-1)^n$, I am not sure what I should do. It would seem to me that the entire thing is undefined, but that is a bad answer.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 28 '18 at 19:31









an4s

2,0892418




2,0892418










asked Apr 28 '18 at 17:59









user25758user25758

306




306












  • $begingroup$
    $ntoinfty$ does not necessarily imply $n$ is an integer, for non integer $n,(-1)^n=?$
    $endgroup$
    – lab bhattacharjee
    Apr 28 '18 at 18:11










  • $begingroup$
    True. Then it would tend to oscillate. I'm not quite sure what I can do with that then. Thanks
    $endgroup$
    – user25758
    Apr 28 '18 at 18:20










  • $begingroup$
    Usually $n$ is used to denote an integer, so assuming it is, note that $cos(-pi)=cos(pi)=-1$.
    $endgroup$
    – Cedron Dawg
    Apr 28 '18 at 19:06




















  • $begingroup$
    $ntoinfty$ does not necessarily imply $n$ is an integer, for non integer $n,(-1)^n=?$
    $endgroup$
    – lab bhattacharjee
    Apr 28 '18 at 18:11










  • $begingroup$
    True. Then it would tend to oscillate. I'm not quite sure what I can do with that then. Thanks
    $endgroup$
    – user25758
    Apr 28 '18 at 18:20










  • $begingroup$
    Usually $n$ is used to denote an integer, so assuming it is, note that $cos(-pi)=cos(pi)=-1$.
    $endgroup$
    – Cedron Dawg
    Apr 28 '18 at 19:06


















$begingroup$
$ntoinfty$ does not necessarily imply $n$ is an integer, for non integer $n,(-1)^n=?$
$endgroup$
– lab bhattacharjee
Apr 28 '18 at 18:11




$begingroup$
$ntoinfty$ does not necessarily imply $n$ is an integer, for non integer $n,(-1)^n=?$
$endgroup$
– lab bhattacharjee
Apr 28 '18 at 18:11












$begingroup$
True. Then it would tend to oscillate. I'm not quite sure what I can do with that then. Thanks
$endgroup$
– user25758
Apr 28 '18 at 18:20




$begingroup$
True. Then it would tend to oscillate. I'm not quite sure what I can do with that then. Thanks
$endgroup$
– user25758
Apr 28 '18 at 18:20












$begingroup$
Usually $n$ is used to denote an integer, so assuming it is, note that $cos(-pi)=cos(pi)=-1$.
$endgroup$
– Cedron Dawg
Apr 28 '18 at 19:06






$begingroup$
Usually $n$ is used to denote an integer, so assuming it is, note that $cos(-pi)=cos(pi)=-1$.
$endgroup$
– Cedron Dawg
Apr 28 '18 at 19:06












1 Answer
1






active

oldest

votes


















1












$begingroup$

Note that $cos$ is even; that is, $cos(x)=cos(-x)$ for any $xinmathbb{R}$. Thus,




  • If $n$ is odd, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft((-1)^{n+1}frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.

  • If $n$ is even, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.


Therefore,
$$lim_{ntoinfty}cosleft((-1)^n frac{n-1}{n+1}piright)=lim_{ntoinfty}cosleft(frac{n-1}{n+1}piright)=cospi=-1.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Simple and clear explanation +1
    $endgroup$
    – Paramanand Singh
    Dec 22 '18 at 7:33











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1












$begingroup$

Note that $cos$ is even; that is, $cos(x)=cos(-x)$ for any $xinmathbb{R}$. Thus,




  • If $n$ is odd, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft((-1)^{n+1}frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.

  • If $n$ is even, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.


Therefore,
$$lim_{ntoinfty}cosleft((-1)^n frac{n-1}{n+1}piright)=lim_{ntoinfty}cosleft(frac{n-1}{n+1}piright)=cospi=-1.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Simple and clear explanation +1
    $endgroup$
    – Paramanand Singh
    Dec 22 '18 at 7:33
















1












$begingroup$

Note that $cos$ is even; that is, $cos(x)=cos(-x)$ for any $xinmathbb{R}$. Thus,




  • If $n$ is odd, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft((-1)^{n+1}frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.

  • If $n$ is even, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.


Therefore,
$$lim_{ntoinfty}cosleft((-1)^n frac{n-1}{n+1}piright)=lim_{ntoinfty}cosleft(frac{n-1}{n+1}piright)=cospi=-1.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Simple and clear explanation +1
    $endgroup$
    – Paramanand Singh
    Dec 22 '18 at 7:33














1












1








1





$begingroup$

Note that $cos$ is even; that is, $cos(x)=cos(-x)$ for any $xinmathbb{R}$. Thus,




  • If $n$ is odd, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft((-1)^{n+1}frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.

  • If $n$ is even, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.


Therefore,
$$lim_{ntoinfty}cosleft((-1)^n frac{n-1}{n+1}piright)=lim_{ntoinfty}cosleft(frac{n-1}{n+1}piright)=cospi=-1.$$






share|cite|improve this answer









$endgroup$



Note that $cos$ is even; that is, $cos(x)=cos(-x)$ for any $xinmathbb{R}$. Thus,




  • If $n$ is odd, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft((-1)^{n+1}frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.

  • If $n$ is even, then $cosleft((-1)^n frac{n-1}{n+1}piright)=cosleft(frac{n-1}{n+1}piright)$.


Therefore,
$$lim_{ntoinfty}cosleft((-1)^n frac{n-1}{n+1}piright)=lim_{ntoinfty}cosleft(frac{n-1}{n+1}piright)=cospi=-1.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 22 '18 at 7:16









choco_addictedchoco_addicted

8,08261947




8,08261947












  • $begingroup$
    Simple and clear explanation +1
    $endgroup$
    – Paramanand Singh
    Dec 22 '18 at 7:33


















  • $begingroup$
    Simple and clear explanation +1
    $endgroup$
    – Paramanand Singh
    Dec 22 '18 at 7:33
















$begingroup$
Simple and clear explanation +1
$endgroup$
– Paramanand Singh
Dec 22 '18 at 7:33




$begingroup$
Simple and clear explanation +1
$endgroup$
– Paramanand Singh
Dec 22 '18 at 7:33


















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