daniton Posted November 6, 2012 Posted November 6, 2012 (edited) Lim[sin(x)/x] as x ->0 Note-it is floor function Edited November 6, 2012 by daniton
daniton Posted November 8, 2012 Author Posted November 8, 2012 i have two things in my mind one the limit is 1 by using the property of limit in combination functions and since floor function is continuous at 1 that is the limit of sin(x)\x the other is when using squeezing theorem we say that x is slightly greater than sin(x) so the ratio is less than 1so the floor of this is 0. what do you say?????????????
mathematic Posted November 9, 2012 Posted November 9, 2012 i have two things in my mind one the limit is 1 by using the property of limit in combination functions and since floor function is continuous at 1 that is the limit of sin(x)\x the other is when using squeezing theorem we say that x is slightly greater than sin(x) so the ratio is less than 1so the floor of this is 0. what do you say????????????? limit = 0 (your analysis is correct).
mathematic Posted November 9, 2012 Posted November 9, 2012 what about the first one?????? floor function is not continuous at integers. floor(1-x) = 0, floor(1+x) = 1, let x -> 0.
alpha2cen Posted November 9, 2012 Posted November 9, 2012 (edited) why?? This is a graph. From the graph we can see sin(x)/x converses to 1 [sin (x)/x] is this graph. Edited November 10, 2012 by alpha2cen
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