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Math Help - limit tan[x]/x

  1. #1
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    Question limit tan[x]/x

    please help my problem . please ....

    limit tan[x]/x
    x->infinite
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  2. #2
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    Re: limit tan[x]/x

    Since $\tan x$ is surjective onto the set of reals, given any $k \in \mathbb{R}$, in each interval $\left(\dfrac{n\pi}{2},\dfrac{(n+1)\pi}{2}\right)$ , there exists $a_n$ such that $\tan(a_n) = ka_n$.

    $\displaystyle \lim_{n \to \infty} a_n = \infty$

    So, since $\displaystyle \lim_{n \to \infty} \dfrac{\tan(a_n)}{a_n} = k$ you know $\displaystyle \lim_{x \to \infty} \dfrac{\tan x}{x}$ does not exist.

    Edit: made the argument more general.
    Last edited by SlipEternal; April 15th 2014 at 12:48 AM.
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  3. #3
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    Re: limit tan[x]/x

    f(x)=sinx/xcosx
    sinx=1 and xcosx=0 for x=Π/2+2nΠ, n=0,1,2,3,..
    So there is no X st:
    L-ε < f(x) < L+ε for all x>X
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  4. #4
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    Re: limit tan[x]/x

    The two answers are wrong
    ؟؟؟؟؟؟؟؟؟
    Because I thought the two before it came into my mind, but they are wrong
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  5. #5
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    Re: limit tan[x]/x

    Please send proof with the final answer؟
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  6. #6
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    Re: limit tan[x]/x

    The solution I posted is the correct answer to the problem you posted. I even proved it. If you think it is wrong, then you posted the wrong question. Is the correct question supposed to be $\displaystyle \lim_{x \to 0} \dfrac{\tan x}{x}$? Because that is a very different question. You asked the limit as $x \to \infty$, which does not exist.
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  7. #7
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    Re: limit tan[x]/x

    Quote Originally Posted by alirezayi9094 View Post
    please help my problem . please ....

    limit tan[x]/x
    x->infinite
    The limit is undefined.

    You can't talk about a limit in this case. To say the limit doesn’t exist is ambiguous. It could mean divergence, which is not the case here.
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  8. #8
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    Re: limit tan[x]/x

    is ur x greatest integer fn here
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  9. #9
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    Re: limit tan[x]/x

    Quote Originally Posted by prasum View Post
    is ur x greatest integer fn here
    Even if it is the greatest integer function, the limit would still be undefined. However, if it is the tangent of the fractional part of x over x, then that is different. Using the squeeze theorem, you have $0 \le \dfrac{\tan \langle x \rangle}{x} < \dfrac{\tan 1}{x}$

    The limit as $x \to \infty$ of the LHS and RHS of the inequality are both 0, so by the Squeeze Theorem, so is the middle limit.
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  10. #10
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    Re: limit tan[x]/x

    Quote Originally Posted by SlipEternal View Post
    \dfrac{\tan \langle x \rangle}{x} < \dfrac{\tan 1}{x}
    That inequality is not true
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  11. #11
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    Re: limit tan[x]/x

    Quote Originally Posted by Shakarri View Post
    That inequality is not true
    Yes it is... as I said, if $\langle x \rangle$ means the fractional part of $x$, that is exactly what it means.
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