Question

How do you find the limit of `cos(x)/(x - pi/2)` as x approaches pi/2?

Answer 1

There's an interesting trick you can do called L'Hopital's rule, since `cos(pi/2) = 0`, and `pi/2 - pi/2 = 0`, giving the proper `0/0` form.

`color(blue)(lim_(x-> pi"/"2) cosx/(x - pi/2)) = lim_(x-> pi"/"2) (d/(dx)[cosx])/(d/(dx)[x - pi/2])`

`= lim_(x->pi"/"2) (-sinx)/(1)`

`= -sin(pi/2)`

`= color(blue)(-1)`

Indeed, at `pi/2 ~~ 1.57`, `f(x) = -1`:

graph{cosx/(x - pi/2) [-7.023, 7.024, -3.51, 3.513]}