# Math Help - Trig Limit Difficulty

1. ## Trig Limit Difficulty

$\lim_{x \to (\pi/2)^+} \frac{cos(x)}{1-sin(x)}$

I get: $= \frac{0}{0}$ I attempt to apply L'Hospital's rule:

$\lim_{x \to (\pi/2)^+} \frac{-sin(x)}{-cos(x)} = \frac{1}{0}$

L'Hospital's rule is no bueno. So, my next thought was to use the squeeze theorem, but since the numerator and denominator have different vertical boundaries, (and the function shoots off unbounded near $\frac{\pi}{2}$) I can't think of a way to do that. I know intuitively that the limit should be $-\infty$, I just can't get there analytically.

2. ## Re: Trig Limit Difficulty

Originally Posted by beebe
$\lim_{x \to (\pi/2)^+} \frac{cos(x)}{1-sin(x)}$

I get: $= \frac{0}{0}$ I attempt to apply L'Hospital's rule:

$\lim_{x \to (\pi/2)^+} \frac{-sin(x)}{-cos(x)} = \frac{1}{0}$

L'Hospital's rule is no bueno. So, my next thought was to use the squeeze theorem, but since the numerator and denominator have different vertical boundaries, (and the function shoots off unbounded near $\frac{\pi}{2}$) I can't think of a way to do that. I know intuitively that the limit should be $-\infty$, I just can't get there analytically.
Actually, L'Hôpital is bueno . $\lim_{x \to (\pi/2)^+} \frac{-\sin(x)}{-\cos(x)}= \lim_{x \to (\pi/2)^+} \frac{\sin(x)}{\cos(x)}\ \to\ \frac{1}{0^-}\ \to\ -\infty$

You can alternatively multiply by the initial rational expression by $\frac{1+\sin(x)}{1+\sin(x)}$ to get

$\frac{\cos(x)}{1-\sin(x)}\frac{1+\sin(x)}{1+\sin(x)}=\frac{\cos(x)( 1+\sin(x))}{1-\sin^2(x)}=\frac{\cos(x)(1+\sin(x))}{\cos^2(x)}$