`lim_(n->00)(n+1)/n=`

Evaluate: lim_(n->oo)(((n+1)(n+2)...(n+n))^(1/n))/n

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

f'(0) = lim_(n->oo) nf(1/n) and f(0)=0 Using this, find lim_(n->oo)((n+1)(2/pi)cos^(- 1)(1/n)-n)),|cos^(-1)1/n|

The value of lim_(n->oo) n^(1/n)

lim_(n->oo)((n^2-n+1)/(n^2-n-1))^(n(n-1)) is

Evaluate: lim_(n->oo)[1/(n a)+1/(n a+1)+1/(n a+2)++1/(n b)]

If A=[{:(,1,a),(,0,1):}] then find lim_(n-oo) (1)/(n)A^(n)

Prove that lim_(n->oo)(1+1/n)^n=e

Evaluate lim_(n->oo)n[1/((n+1)(n+2))+1/((n+2)(n+4))+....+1/(6n^2)]

lim_(n -> oo) (((n+1)(n+2)(n+3).......2n) / n^(2n))^(1/n) is equal to