How do you graph #y=(sinx)/x#?

1 Answer
Apr 19, 2015

We know that the limit in 0 is 1

(it's one of the notables limits: in a neighbourhood of 0 #sin(x)=x+o(x^2) => sin(x)/x = 1+o(x) -> 1 if x->0# )

We know it is an even function (quotient of two odd functions), so the graph must be symmetric.

We concentrate on #x>0#, and then extend by symmetry

We know it has zeros where #sin(x)# has zeros (except for #x=0#) so it has zeros in #x=kpi, k != 0#.

Then we know that #sin(pi/2+2kpi)=1#, so we know that the function in that points is like #1/x#

For the same reason in #x={3pi}/2 + 2kpi# it is like #-1/x#

So you draws the four branches of hyperboles and consider the incidence in the points, consider the zeros, consider than in 0 is 1 and consider the symmetry.

graph{1/x [-10, 10, -5, 5]}
graph{-1/x [-10, 10, -5, 5]}
graph{sinx/x [-10, 10, -5, 5]}