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lim (sinx) ^ 1/x ; x is tending to 0 , x>0. how to solve this limit. plz give the full soluntion
its simple sin x/x ,,,series of sinx is x-[(x^3)/6] taking x common u get 1-[(x^2)/2] and now x tends to 0 so x^2 is negligible
and ans is 1 or
use graph at x→0 (nearly zero) y=x and y= sinx takes same values as y=x is tangent to y=sinx at x=0
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