$r=sec\theta$ ve $y=tan\theta$
olduğundan, $r^2-y^2=\dfrac{1-sin^2\theta}{cos^2\theta}$ olur ve,
$\lim\limits_{\theta \to (\pi/2)^-}\dfrac{1-sin^2\theta}{cos^2\theta}=0/0$ , $cos^2\theta=1-sin^2\theta$ olduğundan limit $1$ dir ve kesin olsun diye l'hôpital alalım,
$\lim\limits_{\theta \to (\pi/2)^-}\dfrac{1-sin^2\theta}{cos^2\theta}=0/0\quad\quad \quad \underbrace{\Longrightarrow}_{l'hôpital uygularız}\quad\quad \quad\lim\limits_{\theta \to (\pi/2)^-}\dfrac{-2sin\theta.cos\theta}{-2cos\theta.sin\theta}=1$