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

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

Evaluate : lim_(n to oo)[(sqrt(n))/((3+4sqrt(n))^(2))+(sqrt(n))/(sqrt(2)(3sqrt(2)+4sqrt(n))^(2))+(sqrt(n))/(sqrt(3)(3sqrt(3)+4sqrt(n))^(2))+.......+(1)/(49n)]

lim_(n rarr oo)(3+sqrt(n))/(sqrt(n))

lim_(n->oo)[1/sqrt(2n-1^2) +1/sqrt(4n-2^2)+1/sqrt(6n-3^2)+...+1/n]

sum_(r=1)^nsin^(-1)((sqrt(r)-sqrt(r-1))/(sqrt(r(r+1)))) is equal to (a) tan^(-1)(sqrt(n))-pi/4 (b) tan^(-1)(sqrt(n+1))-pi/4 (c) tan^(-1)(sqrt(n)) (d) tan^(-1)(sqrt(n)+1)

lim_(n to oo)(1)/(n)(1+sqrt((n)/(n+1))+sqrt((n)/(n+2))+....+sqrt((n)/(4n-3))) is equal to:

underset(nrarroo)lim[(1)/(sqrtn)+(1)/(sqrt(2n))+(1)/(sqrt(3n))+...+(1)/(n)]

The value of lim_(nto oo)(sqrt(n^(2)+n+1)-[sqrt(n^(2)+n+1)]) where [.] denotes the greatest integer function is

If f (theta)=4/3 (1- cos ^(6) theta - sin ^(6)theta), then lim _(ntooo) 1/n [sqrt(f ((1)/(n)))+sqrt(f ((2)/(n)))+sqrt(f((n)/(n)))]=

Evaluate: ("lim")_(n rarr oo)(1/(sqrt(4n^2-1))+1/(sqrt(4n^2-2^2))++1/(sqrt(3n^2)))