For each t∈R, let [t] be the greatest integer less than or equal to t. Then limx→0+ x([1/x] + [2/x] + ..... + [15/x]) is equal to 120. Does not exist (in R). Is equal to 0. Is equal to 15.

L=limx→0+ x[(1x)+...+(15x)]L=limx→0+ x(1x)+x(2x)++x(15x)Takelimx→0+ x[1x]let1x=ywhenx→0,y→∞L=limy→∞ 1yyLety=n+p, whenn→∞,0<p<1limn→∞nn+plimn→∞11+pn=1Takelimx→0+ x[2x]Let2x=y,whenx→0,y→∞,limy→∞ 2yy=2limx→0+ x[1x]+x[2x]+−−−−+x[15x]=1+2+3+−−−−+15=n(n+1)2=15×162=120