solve limits?

Question

prove that Lt x->0+ [sinx /x] is 0
and Lt x->0+ [tanx /x] is 1 where [.] represents Greatest integer function.

Answer 1

this you do by using lhopitals rule which jsut says that whenever you have infinity over infinity, zero over zero (there are other cases) you can just take the derivative of top and bottom, as you know sinx/x the derivative of sin x= cosx and for x is just 1 so you get cosx / 1 at this point you can plug the value for x, that is 0. and you will get cos0/1 = 1... the tanx/x is the same idea since tanx = 0 since sin0/ cos 0 = 0 and x =0 LETS use lHopitials again, oh man this is fun... so derivative of tanx = sec^2x/1 as you know again as x--->0 now you get 1hope it helps

Answer 2

For small positive x the expression x - sin x is positive since its derivative is positive (1 - cos x). So x > sin x for small positive x. So the greatest integer in sin x / x is 0.

Similarly for small positive x the expression tan x - x is positive since its derivative is sec^2 x - 1. Also when x is small and positive then x and tan x are close to one another. Therefore for small positive x the greatest integer in tan x / x is 1.

Answer 3

Use L'Hospital's Rule. Both limits are 1.