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  1. MHB Apprentice

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    #1
    How to solve $ \displaystyle \lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos{x}}{x-\frac{\pi}{2}}$? At first I tried to convert cos x to $ \displaystyle \frac{\tan{x}}{\sin{x}}$ but then realized that $ \displaystyle \lim_{x\rightarrow c}\frac{\tan{x}}{x}$ only applies if c = 0. So, how?

  2. Pessimist Singularitarian
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    #2
    Let's use the substitution:

    $ \displaystyle u=\frac{\pi}{2}-x$

    What does the limit become then?

  3. MHB Apprentice

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    #3 Thread Author
    $ \displaystyle \lim_{u\rightarrow0}\frac{\cos{(\frac{\pi}{2}-u)}}{-u}$?

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    #4
    Quote Originally Posted by Monoxdifly View Post
    $ \displaystyle \lim_{u\rightarrow0}\frac{\cos{(\frac{\pi}{2}-u)}}{-u}$?
    Yes, and so I would factor the negative sign in the denominator out in front of the limit, and apply a co-function identity to the numerator...

  5. MHB Apprentice

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    #5 Thread Author
    Won't it give $ \displaystyle \frac{0}{0}$ again in the $ \displaystyle \frac{\cos{\frac{\pi}{2}}\cdot\cos{u}}{u}$ part?

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    #6
    Quote Originally Posted by Monoxdifly View Post
    Won't it give $ \displaystyle \frac{0}{0}$ again in the $ \displaystyle \frac{\cos{\frac{\pi}{2}}\cdot\cos{u}}{u}$ part?
    Recall the co-function identity:

    $ \displaystyle \cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$

    And so our limit becomes:

    $ \displaystyle L=-\lim_{u\to0}\left(\frac{\sin(u)}{u}\right)$

    And yes, this is an indeterminate form, but it is well-known to students and usually allowed as a result to use:

    $ \displaystyle \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1$

  7. MHB Apprentice

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    #7 Thread Author
    Oops... Sorry sorry... Totally forgot about that...

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    #8
    Consider the following diagram of the unit circle in the first quadrant:



    From this, we can see that for $0\le x\le\dfrac{\pi}{2}$ we have:

    $ \displaystyle \sin(x)\le x\le\tan(x)$

    Dividing though by $\sin(x)$, we obtain:

    $ \displaystyle 1\le\frac{x}{\sin(x)}\le\frac{1}{\cos(x)}$

    or:

    $ \displaystyle \cos(x)\le\frac{\sin(x)}{x}\le1$

    And so we must then have:

    $ \displaystyle \lim_{x\to0}\left(\cos(x)\right)\le\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)\le\lim_{x\to0}(1)$

    or:

    $ \displaystyle 1\le\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)\le1$

    Hence (by the squeeze theorem):

    $ \displaystyle \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1$

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