Find `lim_(n->oo) n/(n+1)`

lim_(n->oo) nsin(1/n)

If f(n+1)=1/2{f(n)+9/(f(n))},n in N , and f(n)>0 for all n in N , then find lim_(n->oo)f(n)

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

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

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

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

If the value of lim_(n->oo){1/(n+1)+1/(n+2)+.......+1/(6n)} is 'K' then find value of (K - log_e 6)? .

If lim_(n->oo)1/((sin^(-1)x)^("n")+1)=1 ,t h e n find the value of x.

Evaluate the following limit: (lim)_(n->oo)(1/(n^2)+2/(n^2)+3/(n^2)++(n-1)/(n^2\ ))