Fri November 11, 2011 By:

what is the limit of (1-2sin^2 (x/2) ) as x tends to 0. can we use the rule: limit[f(x) - g(x)]= limf(x)-limg(x). pls xplain

Expert Reply
Fri November 11, 2011
Yes, limit[f(x) - g(x)]= limf(x)-limg(x) can be used here  since f(x)=1 is a  constant
and it makes no difference to the constant as far as limits are concerned, the value
of constant will remain simply constant. So, you can use this formula.
And using this we get
Lt. x?0 (1-2sin2(x/2) ) = 1 - Lt. x?0 2sin2(x/2) = 1-0 = 1
As sinx = 0 as x goes to zero.
Thank you
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