To prove that lim x⟶0 [sinx/x]=0 , where [.] denotes greatest integer function .
We must first prove that sin x/x tends to 1 from the values that are
less than 1 as x tends to zero
PROOF : Now as x tends to zero, x can take any of the values tending from right of zero and tending from left side of zero
Case - 1:
First we assume that x tends from right of the zero
Now, If x= 0.001, sin(0.001)/0.001 = 0.999999833
If x=0.0001, sin(0.0001/0.0001) = 0.999999998
Case -2:
Now we assume that x is tending from left of the zero
If x= -0.001, sin(-0.001)/-0.001
= sin(0.001)/0.001=0.999999833
If x=-0.0001, sin(-0.0001)/-0.0001
= sin(0.0001)/0.0001 = 0.999999998
Therefore from the above two cases we can see that sinx/x tends to 1 from the values that are less than 1 as x ends to zero
Hence Proved [lim x⟶0 Sinx/x ]=0 {Since [0.....]=0}
Example:
limx⟶3 ( [x-3] + [3 - x] - 4 )
limx⟶3 ( [x] - 3 + 3 + [-x] -4)
= (3 - 3 + 3 - 4 - 4)
= -9
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