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Math Help - Limits of Trignometric Functions

  1. #1
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    Limits of Trignometric Functions

    Can someone answer this please.

    limit as x approaches pi/2 (cotx / (pi/2 - x))

    Any help is appreciated thanx.
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  2. #2
    o_O
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    Have you learned l'hopital's rule?

    \lim_{x \to \frac{\pi}{2}} \left(\frac{\cot x}{\frac{\pi}{2} - x}\right) = \left[\frac{0}{0}\right] = \lim_{x \to \frac{\pi}{2}} \left[ \frac{(\cot x)'}{\left(\frac{\pi}{2} - x\right)'}\right]
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  3. #3
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    Been a while, but I think:

    Note that if you try to evaluate the limit by substitution, you get 0/0, which is an indeterminate form. To evaluate limits with this indeterminate form, you can use l'Hopital's rule. Take the derivative of the numerator and the denominator and THEN evaluate by substitution.

    Taking the derivative gives:

    \frac{-csc^{2}x}{-1}

    Evaluating this by substitution (with x = pi/2) gives the limit to be 1.
    Last edited by Mathnasium; June 5th 2008 at 08:52 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ballin_sensation View Post
    Can someone answer this please.

    limit as x approaches pi/2 (cotx / (pi/2 - x))

    Any help is appreciated thanx.
    Alternatively use the definiton of the derivative at a point

    f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}

    Now rewriting our limit as

    -\lim_{x\to\frac{\pi}{2}}\frac{\cot(x)-\cot\bigg(\frac{\pi}{2}\bigg)}{x-\frac{\pi}{2}}

    we can do this since \cot\bigg(\frac{\pi}{2}\bigg)=0

    So now wee see this is exactly the derivative at point

    now letting f(x)=\cot(x) and c=\frac{\pi}{2}

    we see this is equal to -f'\bigg(\frac{\pi}{2}\bigg)=\csc^2\bigg(\frac{\pi}  {2}\bigg)=1
    Last edited by Mathstud28; June 5th 2008 at 06:59 PM.
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  5. #5
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    No I never did l'hopital's rule but i understand the concept. Thanx for the help.
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  6. #6
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    \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\cot x}{\dfrac{\pi }{2}-x}=\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\sin \bigg( \dfrac{\pi }{2}-x \bigg)}{\dfrac{\pi }{2}-x}\cdot \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{1}{\sin x}=1.
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  7. #7
    o_O
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    As Krizalid's result yielded, the limit is +1. Mathnasium didn't cancel the negative signs, and MathStud28 needed to consider \frac{\pi}{2} - x = -\left(x - \frac{\pi}{2}\right) in the denominator in order for the definition of a derivative approach to work.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    As Krizalid's result yielded, the limit is +1. Mathnasium didn't cancel the negative signs, and MathStud28 needed to consider \frac{\pi}{2} - x = -\left(x - \frac{\pi}{2}\right) in the denominator in order for the definition of a derivative approach to work.
    Thanks man! I didnt even see that, I copied it down wrong!
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