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

1 Answer
May 18, 2015

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.