Calculus Question: Limit of a Sequence Help

I've just been given an assignment online and I can't figure out for the life of me how to solve this problem involving finding the limit of a sequence. Can someone help me out and show me how to do this type of problem so that I can apply it to the rest of my homework, which is 9 more of these almost identical to this one? Thanks so much, here is the problem:

find the limit of the sequence whose terms are given by:

a sub n = (n^2)*(1 - cos(2.6/ n))

Let x be a real number, n be an integer, and f be a function.
If lim x->infinity f(x) exists and equals L, then lim n->infinity f(n) exists and equals L.

So let's first find lim x->infinity (x^2)*(1 - cos(2.6/x)) by using L'Hopital's rule, simplifying, using L'Hopital's rule again, and simplifying again.

lim x->infinity (x^2)*(1 - cos(2.6/x))
= lim x->infinity (1 - cos(2.6/x))/(1/x^2)
= lim x->infinity (d/dx)(1 - cos(2.6/x))/(d/dx)(1/x^2) using L'Hopital's rule on a 0/0 form
= lim x->infinity ((-2.6/x^2)sin(2.6/x))/(-2/x^3)
= lim x->infinity 1.3xsin(2.6/x)
= lim x->infinity 1.3sin(2.6/x)/(1/x)
= lim x->infinity (d/dx)(1.3sin(2.6/x))/(d/dx)(1/x) using L'Hopital's rule on a 0/0 form
= lim x->infinity (-2.6/x^2)(1.3cos(2.6/x))/(-1/x^2)
= lim x->infinity 3.38cos(2.6/x)
= 3.38cos(0)
= 3.38

It then follows that lim n->infinity (n^2)*(1 - cos(2.6/n)) = 3.38.

Lord bless you today!

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