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

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Answer 1 · 2015-05-18T04:00:43

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.

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