What is [sin(x)]/x when the limit of x goes to infinity?

[GeorgeMalvin]

Is this the only time when [sin (x)]/x does not equal to 1?

 

[Dale]

The limit is 0 in this case as
0<=|sin(x)/x|<=1/x
and 1/x->0 as x->infinity.

 

[funny]

Sinx=x-[x^3]/3!+[x^5]/5!-....
Hence, Sinx/x=1-f(x) with f(x) a series which is divergent
Hence, the limit does not exist.