now greates integer finction lies between x and x -1

so that things

((x^2)/(tanxsinx)) -1) < = [(x^2)/(tanxsinx)] < ((x^2)/(tanxsinx)]

so if u take the limit of right hand part by using l hospital rule as its 0/0 and making tanx as sinx/cosx and taking cosx to numerator and sin^2x in the bottom and on applying l hospital rule twice u get it as 1

so answer lies betwen 0 and 1 .

THE ANSWER IS 0 AS RIGHT HAND SIDE EQUALITY ISNT THERE AND ITS A GIF FUNCTION SO ANSWER IS 0.

L HOSPITAL RULE APPLICATION

ON SIMPLYFYING IT

((X^2 *cosx))/(sin^x)

ON APPLYING L HOSPITAL RULE

=(2Xcosx-x^2sinx)/(sin2x)

= (2cosx-2xsinx-2xsinx-x^2cosx)/(2cos2x)

on substituting limit

u get answer as 1

 YOU NEED TO TAKE THE FIRST POINT INTO ACCOUNT

WHENEVER GIF FUNCTION COMES IT LIES BETWEEN X AND X -1

student helper check that sign of equality should be on the RHS  not on the LHS.THEN THE QUESTION WILL GIVE THE PROBLEM

why i dont get it can u please tell once more i dont get what u are coming to say and yes there is a mistake in the first point gif of                       x-1  < [x] < = x. yeah thats a mistake so the answer should be 1 i think.

replying to abv post.

When inequalities come closer to a whole number, it is better to use series expansions for sinx, cosx, tanx etc. otherwise sandwich theorem is enough. And using expansions, it is clearly seen that the limit is 1-. therefore the limit of the [.] is 0

hey cant we simply use the graphs

clearly limx-o sinx/x =1- ( say 1-h where h tends to 0) and also limx-0 tanx/x = 1+( say 1+h)

so we get int (1-h)/1+h = 0 [int is greatest integer function]