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:

1. $$\sec \theta = \frac{1}{\cos \theta}$$
2. $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

$$\frac{1}{\cos \theta}-\frac{\sin\theta}{\cos\theta} = \frac{1-\sin\theta}{\cos\theta} = \frac{1-\sin^2\theta}{(1+\sin\theta)\cos\theta} = \frac{\cos^2\theta}{\cos\theta}\cdot \frac{1}{1+\sin\theta} = \frac{\cos\theta}{1+\sin\theta} = \frac{0}{1+1}$$

When $$\theta \to \frac{\pi}{2}$$ then $$\cos(\frac{\pi}{2}) = 0$$ and $$\sin(\frac{\pi}{2}) = 1$$

$$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta) = 0$$

• It sounds good to me. Well done! – user0102 Feb 21 at 17:23
• Welcome to Mathematics Stack Exchange. You should say $\lim...\color{red}=0$, not $\approx0$ – J. W. Tanner Feb 21 at 17:35

Your proof is fine. Here's another:$$\sec\theta,\,\tan\theta\to\infty\implies\sec\theta+\tan\theta\to\infty\implies\sec\theta-\tan\theta=\frac{1}{\sec\theta+\tan\theta}\to0.$$
• I had an idea earlier that $\frac{1-\sin\theta}{\cos\theta} = 1-\tan\theta \to 0$, given that $\tan \theta \to 1$, or might this rationale be wrong? – Meilton Feb 21 at 19:27
• @jj_ellison Notice in that idea you replaced $1/\cos\theta$ with $1$. While your equation is false, your two sides have the same limit provided $1/\cos\theta-1\to0$, which it would for $\theta\to0$ but not $\theta\to\pi/2$. In any case, $\sec\theta\to1$ only when $\tan\theta\to0\ne1$. – J.G. Feb 21 at 19:44
$$\lim _{\theta\rightarrow 0}\frac{\left(1-\sin \theta \right)}{\cos \theta }=\lim _{\theta\rightarrow 0}\frac{\frac{d}{d \theta}\left(1-\sin \theta \right)}{\frac{d}{d \theta}\left(\cos \theta \right)}=\lim _{\theta\rightarrow 0}\frac{\left(-\cos \theta \right)}{\left(-\sin \theta \right)}=\lim _{\theta\rightarrow 0}\tan \theta =0$$