`lim_(n->oo)1/2tan(x/2)+1/2^2tan(x/2^2)....+1/2^ntan(x/2^n)` is equal to `lim_(n->oo) sum_(n=1)^n1/(2^n)tan(x/(2^n))`

lim_(n->oo)sin(x/2^n)/(x/2^n)

lim_ (n rarr oo) (1) / (2) tan ((x) / (2)) + (1) / (2 ^ (2)) tan ((x) / (2 ^ (2))). ... + (1) / (2 ^ (n)) tan ((x) / (2 ^ (n))) is equal to lim_ (n rarr oo) sum_ (n = 1) ^ (n) (1 ) / (2 ^ (n)) tan ((x) / (2 ^ (n)))

f(n)=lim_(x->0){(1+sin(x/2))(1+sin(x/2^2)).......(1+sin(x/2^n))}^(1/x) then find lim_(n->oo)f(n)

lim_(x rarr oo) (1+2/n)^(2n)=

Evaluate :lim_(n rarr oo)sum_(r=1)^(n)(1)/((n^(2)+r^(2))^(1/2))

lim_(n rarr oo) 1/n^(3) sum_(k=1)^(n) k^(2)x =

The value of the lim_(n->oo)tan{sum_(r=1)^ntan^(- 1)(1/(2r^2))} is equal to

The value of the lim_(n->oo)tan{sum_(r=1)^ntan^(- 1)(1/(2r^2))} is equal to

Lim_(n to 0) sum_(x + 1)^(n) tan^(-1) (1/(2r^2)) equals

lim_(x rarr 2)sum_(n=1)^9x/(n(n+1)x^2+2(2n+1)x+4)