Question

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

Answer 1

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]}