Thread: L'Hopital rule

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

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:

$\displaystyle
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:

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

RonL

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 $\displaystyle \frac 00$ or $\displaystyle \frac {\infty}{\infty}$

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

Now apply L'Hopital's