Try using the expansion of sinx and you will notice that sinx/x will always be less than 1 which impiles that greatest integer satisfying this is 0.
the value of:
lim [|sin x||x|]
x→0
[.] represents greatest integer function
|.| represents modulus....
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8 Answers
Abhishek Majumdar
·2010-09-24 23:09:37
i got the answr as 1....bt the answer is given 0 in the ARIHANT DIFFERENTIAL CALCULUS book
vishuboy
·2010-09-25 05:15:21
Asish Mahapatra
·2010-09-25 05:24:49
or simply,
consider f(x) = sinx - x
f(0) = 0
f'(x) = cosx - 1 < 0 for all x
=> sinx - x is decreasing for x > 0
=> sinx < x for x > 0
=> |sinx| < |x| for all x ≠0
=> |sinx|/|x| < 1 for all x≠0
hence [|sinx||x|] = 0 for all x ≠0
and since you are calculating limit, lim(x→0) ≠x = 0
there fore x is non-zero hence the limit = 0
rishan chattaraj
·2010-10-01 23:56:45
but if we calculate the limit using the conventional methods,
both the L.H.L.and the R.H.L comes to 0
why the difference?