Question

What is the limit of `(1+sec^3 x)/tan^2 x` as x approaches 180?

Answer 1

I assume that `x` is approaching `180^@` (not radians).
I also assume that you want to find this limit without resorting to l'Hopital's rue.

`1=sec^3x` is a sum of two cubes:

`1=sec^3x = (1+secx)(1-secx+sec^2x)`

We also know that `tan^2x+1=sec^2x`, so `tan^2x = sec^2x-1`.
And `sec^2x-1` is a difference of squares, so we can factor it too.

Rewrite
`(1+sec^3x)/tan^2x = ((1+secx)(1-secx+sec^2x))/((secx+1)(secx-1)) =(1-secx+sec^2x)/(secx-1)`

Now evaluate the limit by substitution.