Question

Does the limit exist? hot to find that? [see attachment]

Answer 1

multiply the denominator

Answer 2

i cant simplify it.

Answer 3

you'll get powers of 2 on the bottom

Answer 4

yea, that's what ive done~ stuck~~ -,-

Answer 5

devide every term by n^2 then run the limit

Answer 6

or reason as follows
we can ignore \(\sin^2(n)\) because it is bounded by 1 and -1
we can ignore the \(\frac{2}{n}\) because it goes to zero as \(n\to \infty\)

Answer 7

removing these terms gives a polynomial of degree 2 top and bottom
take the ratio of the leading coefficients