1. ## L'Hopital rule

limx->90' (x-90')^2secx=??

2. Originally Posted by Joyce
limx->90' (x-90')^2secx=??
Is that meant to be 90 degrees? The first thing to do is rewrite this in radians:

$
L = \lim_{x \to \pi/2} (x-\pi/2)^2 \sec(x) = \lim_{x \to \pi/2} \frac{(x-\pi/2)^2} {\cos(x)}
$

Then if we must use L'Hopital's rule:

$
L = \lim_{x \to \pi/2} \frac{2(x-\pi/2)}{-\sin(x)}=0
$

RonL

3. Originally Posted by Joyce
limx->90' (x-90')^2secx=??
Do you know L'Hopital's rule? to use it, we must have the condition that the limit goes to $\frac 00$ or $\frac {\infty}{\infty}$

so, rewrite $\lim_{x \to 90^{\circ}} \left( x - 90^{\circ} \right)^2 \sec x$ as $\lim_{x \to 90^{\circ}} \frac {\left( x - 90^{\circ} \right)}{\cos x}$

Now apply L'Hopital's