We have to find the the greatest integer part of the value [sinx]. we know that sin pi/2 = 1 but since the limit is approaching pi/2 but is not completely pi/2, therefore we can conclude by the rules of greatest integer function that the value of function greater than 0 and slightly less than 1 is always zero(only if it is the greatest integer function).
At exactly x=1 without limits step of sinx is 1 but limit x tends to means it approaches to not equal to then the value is always less than 1 if we apply step function it is zero
When we consider the value of sin[pi/2] it gets in between 0 and 1. And we know that, [ ] this bracket indicates the greatest integer function. therefore,when we consider the both left hand limit and right hand limit they both equal each other to 0. that's why 0 is the correct answer.
Actually sinπ/2 is 1, but the question is greatest integer function of sinx i.e, [sinx] and the value is not exactly 1 as limit approaches π/2 and not exactly π/2. So it is to be considered as 0 by the greatest integer function rule. So answer is 0. Option C is correct.
In these types of questions we have to consider g.i.f .As sinx is approaching 1 i.e it is less than 1, the answer will be 0.But if the whole limit would have been in the gif the answer would be 1.
As limit x approaches to π\2.now put value in (sinx) from R.H.L then we get + sin π/2 and from L.H. L side we get -sinπ/2. And at 0 we simply get sinπ/2 and we also know that sin 90 is zero so in above all three values become 0 from R. H. L & L. H. L also it is 0 at x= 0.hence
LHL :- h => 0 [Sin(π/2 - h)] = limh => 0 [Cosh] so, Cosh => 1 so, it's become zero. RHL :- h=> 0 [sin ( π/2 + h)] = lim h=>0 [Cosh] here h tending to zero means cosh tending to. 1so it's also zero. so zero answer
Sin(x) tending to 90° means either greater than 90° or less then 90° in both the cases it will less the zero according to it's graph hence gif of any number less the one is absolute zero. good question
The question really mean that if x approach 90*then sinx approach what, yes it approach 1,but not exactly 1,it may be 0.99 something ,you may think it may be 1.01 but sin value doesn't exceed 1.so G.I.F of less than one is zero
Just see limit is in greatest integer function , since range of sinx is -1 to 1 .From graph you can see when you see left hand limit you got a number less than 1 whose greatest integer is zero(0) similarly when you see right hand limit you got a number less than 1 whose greatest integer is zero(0) SINCE RHL=LHL=0 SO LIMIT OF FUNCTION EXISTS AT pi/2 and equal to zero(0)
[sinx] is zero at pi/2+ (at GIF) [sinx]is zero at pi/2-(at LIF) hence , LIF=GIF (it is a continuous) So, zero is the answer.. the figures shows parabolic nature.. ..
Lt sin x .... that is simply take derivative of sin x... derivative of sin x is cos x... now put value of x cos π/2 = 0.. hence correct option is C that is 0
Here as the x tends to pi by 2 function tends to 1 but not equals to 1 so the graph lies between y=0 and 1 in this greatest integer function is 0 Hence answer is 0