For each t∈R, let [t] be the greatest integer less than or equal to t. Then,limx→0+ x1x+2x+……+15x

We know that [x]=x−{x}, where {x} denotes the fractional part of x.∴limx→0+ x1x+2x+…+15x−1x+2x+…+15x=limx→0+ (1+2+…+15)−limx→0+ x1x+2x+…+15x=15(15+1)2+0×A finite number less than 15=120