Lim[sin(x)/x] as x ->0

 

Note-it is floor function

Edited November 6, 201213 yr by daniton

What do you think the limit is? Why?

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?????????????

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).

what about the first one??????

I think it converges to 1 as x → 0.

why??

what about the first one??????

 

 

floor function is not continuous at integers. floor(1-x) = 0, floor(1+x) = 1, let x -> 0.

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, 201213 yr by alpha2cen