computing the limit $\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$
I'm trying to compute the following limit and would greatly appreciate your heartening feedback on my solution.
The limit:
$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$
My steps in deriving the solution:
Preliminary identities:
- $\sec \theta = \frac{1}{\cos \theta}$
- $\tan \theta = \frac{\sin \theta}{\cos \theta}$
$\frac{1}{\cos \theta}-\frac{\sin\theta}{\cos\theta} = \frac{1-\sin\theta}{\cos\theta} = \frac{1-\sin^2\theta}{(1+\sin\theta)\cos\theta} = \frac{\cos^2\theta}{\cos\theta}\cdot \frac{1}{1+\sin\theta} = \frac{\cos\theta}{1+\sin\theta} = \frac{0}{1+1}$
When $\theta \to \frac{\pi}{2}$ then $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$
The answer being:
$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta) = 0$
Your proof is fine. Here's another:$$\sec\theta,\,\tan\theta\to\infty\implies\sec\theta+\tan\theta\to\infty\implies\sec\theta-\tan\theta=\frac{1}{\sec\theta+\tan\theta}\to0.$$
Your approach is indeed the best way to tackle such problems, here is alternative way using L's hospital rule after applying the preliminary identities...
$\lim _{\theta\rightarrow 0}\frac{\left(1-\sin \theta \right)}{\cos \theta }=\lim _{\theta\rightarrow 0}\frac{\frac{d}{d \theta}\left(1-\sin \theta \right)}{\frac{d}{d \theta}\left(\cos \theta \right)}=\lim _{\theta\rightarrow 0}\frac{\left(-\cos \theta \right)}{\left(-\sin \theta \right)}=\lim _{\theta\rightarrow 0}\tan \theta =0$
Again, that was absolutely unnecessary, yet just in case it would be useful later...... :~)