# Intuition behind a certain limit.

We want to find $\displaystyle\lim_{\theta\to\frac{\pi}{2}} b_1-a_1$, we are given $c=1$ and that $\cdot=90^{\circ}$

This is my solution;

$$$$\sin \theta=\frac{b_1}{a_1} \iff b_1=a_1 \sin \theta\tag{1}$$$$

$$$$\tan \theta=\frac{b_1}{c}\iff \tan \theta=b_1\tag{2}$$$$

$$$$b_1=\tan \theta\stackrel{(1)}{\iff}a_1= \sec \theta\tag{3}$$$$

$$\lim_{\theta \to \frac{\pi}{2}} b_1-a_1\stackrel{(2)(3)}{=}\lim_{\theta\to\frac{\pi}{2}}\tan \theta-\sec \theta=-\lim_{\theta\to\frac{\pi}{2}}\frac{\cos \theta \tan \theta -\cos^2\theta\tan^2\theta}{\cos^2\theta\tan \theta}=-\lim_{\theta\to\frac{\pi}{2}}\frac{1-\cos \theta \tan \theta }{\cos \theta}\stackrel{\text{de l'}}{=}-\dfrac{1}{-\infty}=0$$

Assuming I did not do any mistake in there, that means that when the angle tends to become a right angle, $a_1=b_1$. But according to my intuition, $a_1>b_1$ no matter the angle. Where is my mistake?

• Note that $\tan \theta =\frac{b_1}{c}.$
– mfl
Oct 27 '14 at 23:20
• @mfl lol! Editing. Oct 27 '14 at 23:21

## 1 Answer

There is no mistake. It is perfectly possible that $f(\theta) > g(\theta)$ for all $\theta$ but $\lim f(\theta)-g(\theta) = 0$.

To convince yourself maybe you could try to find a simpler proof. The suggestion of @DanielFischer is good: $a^2 = 1+b^2$ means that $$(a+b)(a-b) = 1\qquad a-b = \frac{1}{a+b}$$

• I know that this is possible. However, there has to be a mistake either on my intuitive answer or in the analytical mathematical one. If the math is correct, then I have to spot the mistake in my intuition. Oct 27 '14 at 23:18
• @UserX Does it help your intuition any to take $b_1=1$ and consider the limit as $c\to 0$? (You should be able to show the equivalence of the two problems) Oct 27 '14 at 23:55
• @StevenStadnicki yes. I got it. I can intuitively tell the two problems are a tautology but I don't know how to show it formally. Oct 28 '14 at 0:04