Find the limit of the sequence

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  • #1
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Find the limit of the sequence whose terms are given by [tex]a_n = (n^2)(1-cos(5.2/n))[/tex]

well as n->inf, cos goes to 1 right? so shouldnt the limit of this sequence be 0?
 

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  • #2
saltydog
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ProBasket said:
Find the limit of the sequence whose terms are given by [tex]a_n = (n^2)(1-cos(5.2/n))[/tex]

well as n->inf, cos goes to 1 right? so shouldnt the limit of this sequence be 0?
Mathematica returns:

[tex]\mathop \lim\limits_{n\to \infty}n^2[1-Cos(\frac{a}{n})]=\frac{a^2}{2}[/tex]

I'd like to know how too.
 
  • #3
shmoe
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But [tex]n^2[/tex] goes to infinity, so your limit is an indeterminate form, infinity*0. Rewrite as:

[tex]a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}[/tex]

and use l'hopital, or you could use the taylor series for cos.
 
  • #4
saltydog
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shmoe said:
But [tex]n^2[/tex] goes to infinity, so your limit is an indeterminate form, infinity*0. Rewrite as:

[tex]a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}[/tex]

and use l'hopital, or you could use the taylor series for cos.
Right, just twice. Thanks.
 

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