Evaluate :
`lim_( n -> oo ) Sin( a/2^n ) / Sin( b/2^n )`

lim_(n->oo)(sin((a)/(2^n)))/(sin((b)/(2^(n+1))))=

Evaluate lim_ (n rarr oo) n ^ (2) int _ (- (1) / (2)) ^ ((1) / (pi)) (2006sin x + 2007cos x) | x | dx

Evaluate: lim_ (n rarr oo) (1 ^ (4) + 2 ^ (4) + 3 ^ (4) + ... + n ^ (4)) / (n ^ (5)) - lim_ (n rarr oo) (1 ^ (3) + 2 ^ (3) + ... + n ^ (3)) / (n ^ (5))

Evaluate: lim_ (n rarr oo) (1 ^ (4) + 2 ^ (4) + 3 ^ (4) + ... + n ^ (4)) / (n ^ (5)) - lim_ (n rarr oo) (1 ^ (3) + 2 ^ (3) + ... + n ^ (3)) / (n ^ (5))

lim_ (n rarr oo) (2 ^ (n) -1) / (2 ^ (n) +1)

lim_ (n rarr oo) (2 ^ (n) -1) / (2 ^ (n) +1)

lim_(n rarr oo)2^(n)sin(a)/(2^(n))

Evaluate: lim_(n rarr oo)(n^(p)sin^(2)(n!))/(n+1)

The value of lim_( n to oo) (1)/(n) ( sin (pi/ n) + sin (2pi/n) + …+ sin (n pi/ n) ) is