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Math Help - L'Hopital rule

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    L'Hopital rule

    limx->90' (x-90')^2secx=??
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  2. #2
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    Quote Originally Posted by Joyce View Post
    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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Joyce View Post
    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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