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Thread: The limit as x approaces zero of cscx - 1/x

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    The limit as x approaces zero of cscx - 1/x

    Problem 29 from chapter 9, section 2 of the 5th edition of Calculus with Analytic Geometry by Purcell and Varberg.

    My answer disagrees with the answer key, which gives $\infty$. The problem has the form $\infty - \infty$. The second and third equalities come from l'H˘pital's rule, which I applied because the second and third items have the form $\frac{0}{0}$.

    $\displaystyle \lim_{x \to 0}\left(\csc x - \frac{1}{x}\right) = \lim_{x \to 0}\frac{x - \sin x}{x \sin x} = \lim_{x \to 0}\frac{1 - \cos x}{x \cos x + \sin x} = \lim_{x \to 0}\frac{\sin x}{2\cos x - x\sin x} = 0$
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    Re: The limit as x approaces zero of cscx - 1/x

    Lookin at a graph, the limit is clearly 0.
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    Re: The limit as x approaches zero of cscx - 1/x

    That's cheating, but both https://www.desmos.com/calculator and numpy agree. That makes three answer key errors in just this section. It had $0$ for $\displaystyle\lim_{n \to \infty}\sqrt[n]{a}$ and $\displaystyle\lim_{n \to \infty}\sqrt[n]{n}$. Still very rare overall.
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    Re: The limit as x approaches zero of cscx - 1/x

    Quote Originally Posted by Zexuo View Post
    That's cheating, but both https://www.desmos.com/calculator and numpy agree. That makes three answer key errors in just this section. It had $0$ for $\displaystyle\lim_{n \to \infty}\sqrt[n]{a}$ and $\displaystyle\lim_{n \to \infty}\sqrt[n]{n}$. Still very rare overall.
    THAT is true nonsense

    Look at THIS
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    Re: The limit as x approaces zero of cscx - 1/x

    \begin{align}\lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) &= \lim_{x \to 0} \frac{x - \sin{(x)}}{x \sin{(x)}} = \lim_{x \to 0} \frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \lim_{x \to 0} \left( \frac{2\cos{(x)}}{2\cos{(x)}} \cdot \frac{x - \sin{(x)}}{x \sin{(x)}} \right) \\ &= \lim_{x \to 0}\frac{2x\cos{(x)} - 2\sin{(x)}\cos{(x)}}{2x \sin{(x)}\cos{(x)}} \\ &= \lim_{x \to 0}\frac{2x\cos{(x)} - \sin{(2x)}}{x \sin{(2x)}} \\ &= 2\lim_{x \to 0}\frac{2x(\cos{(x)}-1)+2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= 2\lim_{x \to 0}\frac{2x(\cos{(x)}-1)}{2x \sin{(2x)}}+2\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \lim_{x \to 0}\frac{2x(\cos{(x)}-1)}{x \sin{(2x)}}+2\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \lim_{x \to 0} \left(\frac{2x}{\sin{(2x)}} \cdot \frac{\cos{(x)}-1}{x}\right)+2\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \cancelto{1}{\left(\lim_{x \to 0} \frac{2x}{\sin{(2x)}}\right)} \cdot \cancelto{0}{\left(\lim_{x \to 0} \frac{\cos{(x)}-1}{x}\right)}+2\cancelto{\lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right)}{\left(\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}}\right)} \\ \lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) &= 2\lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) \\ \lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) &= 0
    \end{align}
    Of course, the final step is only valid if the limit exists.
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    Re: The limit as x approaches zero of cscx - 1/x

    Quote Originally Posted by Plato View Post
    THAT is true nonsense

    Look at THIS
    Only if you are allowed a calculator to do the problem...

    -Dan
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    Re: The limit as x approaches zero of cscx - 1/x

    Quote Originally Posted by topsquark View Post
    Only if you are allowed a calculator to do the problem...
    Have a look at this opinion piece from the mathguy on public radio.
    Then think about this question I wrote in response.
    The limit as x approaces zero of cscx - 1/x-testii.gif
    Given the graph of $f(x)$ which of the following is the largest value?
    $\begin{cases}a)~\int_0^6 {f(x)dx} &~b)\int_0^6 {f^2(x)dx} \\
    c)~\int_0^6 {\sqrt{f(x)}dx} &d)~\int_2^6 {f^3(x)dx}\\e)~\text{cannot be determined}\end{cases}$

    Over a period of years I gave this in a talk at to several meetings.
    Meeting of secondary teachers of AP calculus & community college mathematics teachers.
    In all of those sessions, I had only one correct answer.
    I summit that this is exactly what Devlin was about in that link.
    No calculator, computer , or webpage would be of any help!
    Do you know the answer? Any one?
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    Re: The limit as x approaches zero of cscx - 1/x

    Quote Originally Posted by Plato View Post
    Have a look at this opinion piece from the mathguy on public radio.
    Then think about this question I wrote in response.
    Click image for larger version. 

Name:	testii.gif 
Views:	6 
Size:	3.5 KB 
ID:	39170
    Given the graph of $f(x)$ which of the following is the largest value?
    $\begin{cases}a)~\int_0^6 {f(x)dx} &~b)\int_0^6 {f^2(x)dx} \\
    c)~\int_0^6 {\sqrt{f(x)}dx} &d)~\int_2^6 {f^3(x)dx}\\e)~\text{cannot be determined}\end{cases}$

    Over a period of years I gave this in a talk at to several meetings.
    Meeting of secondary teachers of AP calculus & community college mathematics teachers.
    In all of those sessions, I had only one correct answer.
    I summit that this is exactly what Devlin was about in that link.
    No calculator, computer , or webpage would be of any help!
    Do you know the answer? Any one?
    Given that on $\displaystyle 0<y<1$, we have $\displaystyle y^a>y>y^b$ where $\displaystyle a<1<b$, I would say that c) is the correct choice here.
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    Re: The limit as x approaches zero of cscx - 1/x

    Quote Originally Posted by MarkFL View Post
    Given that on $\displaystyle 0<y<1$, we have $\displaystyle y^a>y>y^b$ where $\displaystyle a<1<b$, I would say that c) is the correct choice here.
    I independently agree with you and for the same reason. Though I admit there was a Physics-y version I flunked once before so I have had some little experience with this kind of "paradox."

    -Dan
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    Re: The limit as x approaches zero of cscx - 1/x

    Sin(x) is an odd function, and it can be approximated by $ \ (x + kx^3) \ $
    for x relatively close to 0 and for k equal to a real number.


    $\displaystyle \lim_{x \to 0}\left(\csc(x) - \frac{1}{x}\right) \ = $

    $\displaystyle \lim_{x \to 0}\left(\frac{1}{sin(x)} - \frac{1}{x}\right) \ \approx \ $

    $\displaystyle \lim_{x \to 0}\left(\frac{1}{x + kx^3} - \frac{1}{x}\right) \ = \ $

    $\displaystyle \lim_{x \to 0}\left(\frac{1}{x + kx^3} - \frac{1(1 + kx^2)}{x(1 + kx^2)}\right) \ = \ $

    $\displaystyle \lim_{x \to 0}\left(\frac{1 - 1 - kx^2}{x + kx^3} \right) \ = \ $

    $\displaystyle \lim_{x \to 0}\left(\frac{x(- kx)}{x(1 + kx^2)} \right) \ = \ $

    $\displaystyle \lim_{x \to 0}\left(\frac{- kx}{1 + kx^2} \right) \ = \ $

    0
    Last edited by greg1313; Dec 17th 2018 at 02:00 PM.
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