lim n→∞ [x]+[2x]+[3x]+.........+[nx]

---

ⁿ²

                                  ⁿ²

is =

lim n→∞  x+2x+3x+.........+nx + {x}+{2x} ...+{nx}

= x(1+2+3+..+n)   +   {x}+...{nx}

=x(n(n+1))/2n^2 +  ({x}+...{nx})/n^2

Since {} is only b/w 0 and 1  , the second operand becomes 0 as n tends to ∞

on solving the first part , u get (n² x +nx)/2n² = x/2 +x/2n

x/2n becomes 0 as x tends to infinity.

therefore the answer is x/2.

An easy way to do limits when Greatest integer  function is there, is to take out the commom element which does not change ... in this case x. (x , 2x, 3x ...)  x remains same.

divide this element by the power to which the denominator (having the dependent variable) is raised.

eg: [x]+[2x]+[3x]....+[nx]

as n tends to infinity, = x/3