What is the Limit of the Sequence a_n = (n^2)(1-cos(5.2/n))?

[ProBasket]

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

well as n->inf, cos goes to 1 right? so shouldn't the limit of this sequence be 0?

 

[saltydog]

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

well as n->inf, cos goes to 1 right? so shouldn't the limit of this sequence be 0?

Mathematica returns:

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

I'd like to know how too.

 

[shmoe]

But n^2 goes to infinity, so your limit is an indeterminate form, infinity*0. Rewrite as:

a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}

and use l'hopital, or you could use the taylor series for cos.

 

[saltydog]

But n^2 goes to infinity, so your limit is an indeterminate form, infinity*0. Rewrite as:

a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}

and use l'hopital, or you could use the taylor series for cos.

Right, just twice. Thanks.