[" For each "_(t in R),[t]" denote greatest integer less than or equal to "t" then "Lim_(t rarr pi)x[(1)/(x)]+[(2)/(x)]+...+[(15)/(x)])" is "],[" equal to "]

For each tinR," let "[t] be the greatest integer less than or equal to t. Then lim_(xto0^(+)) x([(1)/(x)]+[(2)/(x)]+...+[(15)/(x)])

For each t in R let [t] be the greatest integer less than or equal to t then lim_(x rarr0^(+))x([(1)/(t)]+[(2)/(t)]+...+[(15)/(t)])(1) is equal to 0(2) is equal to 15(3) is equal to 120 (4) does not exist (in R)

For each tinR," let "[t] be the greatest integer less than or equal to t. Then find lim_(xto0^(+)) x([(1)/(x)]+[(2)/(x)]+...+[(15)/(x)])

if [x] denotes the greatest integer less than or equal to x, than lim_(xrarr0)(x[x])/(sin|x|) , is

If n in N , and [x] denotes the greatest integer less than or equal to x, then lim_(xrarrn)(-1)^([x]) is equal to.

If [x] denotes the greatest integer less than or equal to x, then the value of lim_(x rarr1)(1-x+[x-1]+[1-x]) is

For each t in RR, let [t] be the greatest integer less than or equal to t. Then, lim_(x rarr 1+) ((1-|x|+sin|1-x|)sin(pi/2[1-x]))/(|1-x|[1-x])

For each t in R ,let[t]be the greatest integer less than or equal to t. Then lim_(xto1^+)((1-absx+sinabs(1-x))sin(pi/2[1-x]))/(abs(1-x)[1-x])

For each t in R ,let[t]be the greatest integer less than or equal to t. Then lim_(xto1^+)((1-absx+sinabs(1-x))sin(pi/2[1-x]))/(abs(1-x)[1-x])