`lim_(x -> 2) (sum_(r = 1) ^n x^r - sum_(r = 1) ^n 2^r) / (x - 2)`

lim_(xrarr2)(sum_(r=1)^(n)x^r-sum_(r=1)^(n)2r)/(x-2) is equal to

lim_(xrarr2)(sum_(r=1)^(n)x^r-sum_(r=1)^(n)2^r)/(x-2) is equal to

lim_ (x rarr oo) sum_ (r = 0) ^ (n) ((1) / ((r + 2) n ^ (r)) =

lim_ (n rarr oo) (1) / (n ^ (4)) sum_ (r = 1) ^ (n) r (r + 2) (r + 4) =

Let f (x) = lim_ (n rarr oo) sum_ (r = 0) ^ (n-1) (x) / ((rx + 1) {(r + 1) x + 1})

lim_(xto1) (sum_(r=1)^(n)x^(r)-n)/(x-1) is equal to

If alpha and beta are the roots of the equation 375x^(2)-25x-2=0, then the value of lim_(n rarr oo)(sum_(r=1)^(n)alpha^(r)+sum_(r=1)^(n)beta^(r)) is

If f(x) = lim_(n->oo) sum_(r=0)^n (tan(x/2^(r+1)) + tan^3 (x/2^(r+1)))/(1- tan^2 (x/2^(r+1))) then lim_(x->0) f(x)/x is

If f(x) = lim_(n->oo) sum_(r=0)^n (tan(x/2^(r+1)) + tan^3 (x/2^(r+1)))/(1- tan^2 (x/2^(r+1))) then lim_(x->0) f(x)/x is