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 $\sin(\frac{\pi}{2}) = 1$

The answer being:

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

  • 8
    $\begingroup$ It sounds good to me. Well done! $\endgroup$
    – user0102
    Feb 21, 2021 at 17:23
  • $\begingroup$ Welcome to Mathematics Stack Exchange. You should say $\lim...\color{red}=0$, not $\approx0$ $\endgroup$ Feb 21, 2021 at 17:35

2 Answers 2


Your proof is fine. Here's another:$$\sec\theta,\,\tan\theta\to\infty\implies\sec\theta+\tan\theta\to\infty\implies\sec\theta-\tan\theta=\frac{1}{\sec\theta+\tan\theta}\to0.$$

  • $\begingroup$ 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? $\endgroup$
    – Meilton
    Feb 21, 2021 at 19:27
  • $\begingroup$ @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 $\tan\theta\to0\ne1$. $\endgroup$
    – J.G.
    Feb 21, 2021 at 19:44
  • $\begingroup$ I have not seen your answer, I have put another and now I have seen that it is the same of your. +1 $\endgroup$
    – Sebastiano
    Feb 21, 2021 at 21:12

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...... :~)


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