L'Hopital rule

**Joyce** · Oct 5th 2007, 06:26 AM

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

**CaptainBlack** · Oct 5th 2007, 06:49 AM

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:

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

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

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

RonL

**Jhevon** · Oct 5th 2007, 06:50 AM

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

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