Question

How do you find `lim costheta/(pi/2-theta)` as `theta->pi/2` using l'Hospital's Rule?

Answer 1

`1`

Explanation:

l'Hospital's Rule states that if `lim_(x->c) f(x)/g(x) = 0/0` aka. `f(c) = 0` and `g(c) = 0`, then the limit can be written as `lim_(x->c) (f'(x))/(g'(x))`. If you are still in indeterminate form (`0/0`), then you use l'Hospital's Rule again.

`lim_(theta -> pi/2) (costheta)/(pi/2 - theta) = lim_(theta -> pi/2)(-sintheta)/(-1) = lim_(theta -> pi/2) sin theta = sin(pi/2) = 1`

Hopefully this helps!