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Math Help - L'Hopital's Rule

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question L'Hopital's Rule

    the limit of (x tan (1/x)) as x aproaches 0 from the right side

    Im sorry, I dont know how to put it in LaTex Fomat :/

    My first thought is to place x as x^-1 and take the derivative of tan \frac {1}{x} but it doesn't seem right.
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  2. #2
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    \lim_{x\to 0^{+}}xtan(\frac{1}{x})

    This one does not appear to exist. Look at the graph and you can see.
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by galactus View Post
    \lim_{x\to 0^{+}}xtan(\frac{1}{x})

    This one does not appear to exist. Look at the graph and you can see.
    in the book it says the answer is 1...but i don't see how they approached the problem...
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  4. #4
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    in the book it says the answer is 1...but i don't see how they approached the problem...
    wait, I Am Very Sorry, it is the limit as it approaches infinity
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  5. #5
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    Here is a graph of it. I do not know where they got 1. It looks undefined to me.
    Attached Thumbnails Attached Thumbnails L'Hopital's Rule-tan.jpg  
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  6. #6
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by galactus View Post
    Here is a graph of it. I do not know where they got 1. It looks undefined to me.
    its an error, it aproaches infinity, does that change the problem?
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  7. #7
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    wait, I Am Very Sorry, it is the limit as it approaches infinity
    OK. That's different then.

    \lim_{x\to {\infty}}xtan(\frac{1}{x})

    L'Hopital:

    \lim_{x\to \infty}\left(1+tan^{2}(\frac{1}{x})\right)

    =1+\left(\lim_{x\to {\infty}}tan(\frac{1}{x})\right)^{2}

    =1+tan^{2}\left(\frac{1}{\lim_{x\to {\infty}}}\right)=1
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  8. #8
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    wait, I Am Very Sorry, it is the limit as it approaches infinity
    \lim_{x \to \infty} \frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}}

    \lim_{x \to \infty} \frac{\sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}

    \lim_{x \to \infty} \sec^2\left(\frac{1}{x}\right) = 1
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  9. #9
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by galactus View Post
    OK. That's different then.

    \lim_{x\to {\infty}}xtan(\frac{1}{x})

    L'Hopital:

    \lim_{x\to \infty}\left(1+tan^{2}(\frac{1}{x})\right)

    =1+\left(\lim_{x\to {\infty}}tan(\frac{1}{x})\right)^{2}

    =1+tan^{2}\left(\frac{1}{\lim_{x\to {\infty}}}\right)=1
    How did you tan^2? I got lost after that :/

    Quote Originally Posted by skeeter View Post
    \lim_{x \to \infty} \frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}}

    \lim_{x \to \infty} \frac{\sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}

    \lim_{x \to \infty} \sec^2\left(\frac{1}{x}\right) = 1
    I was doing just like how you were doing it, but i got stuck in the last part: sec^2 (1/x)

    to make it clear, is (1/x) as x approaches infinity, is 0? what if it was approaches 0? [just 1/x]
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  10. #10
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    sec^{2}(u)=1+tan^{2}(u)

    It's the same thing only different
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  11. #11
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by galactus View Post
    sec^{2}(u)=1+tan^{2}(u)

    It's the same thing only different
    wait, then wouldn't it be 1 + 1 + tan^2? what happened to x?
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  12. #12
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    xtan(\frac{1}{x}) = x\cdot \frac{sin(\frac{1}{x})}{cos(\frac{1}{x})} = \frac{ \frac{sin(\frac{1}{x})}{\frac{1}{x}}}{cos(\frac{1}  {x})}

    Now,

    \lim_{x \to \infty} \frac{ \frac{sin(\frac{1}{x})}{\frac{1}{x}}}{cos(\frac{1}  {x})} \overbrace{=}^{u = \frac{1}{x}} \lim_{u \to 0} \frac{sin(u)}{u} \cdot \frac{1}{cos(u)} = \lim_{u \to 0}\frac{sin(u)}{u}\cdot \lim_{u \to 0}\frac{1}{cos (u)} = 1 \cdot 1 = 1
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  13. #13
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    Hello, >_<SHY_GUY>_<!

    \lim_{x\to\infty}x\tan\frac{1}{x}
    Let u = \frac{1}{x}

    Then we have: . \lim_{u\to0}\left(\frac{1}{u}\cdot\tan u\right) \;=\;\lim_{u\to0}\left(\frac{1}{u}\cdot\frac{\sin u}{\cos u} \right)\;=\;\lim_{u\to0}\left(\frac{\sin u}{u}\cdot\frac{1}{\cos u}\right) \;=\;1\cdot\tfrac{1}{1} \;=\;1

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  14. #14
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, >_<SHY_GUY>_<!

    Let u = \frac{1}{x}

    Then we have: . \lim_{u\to0}\left(\frac{1}{u}\cdot\tan u\right) \;=\;\lim_{u\to0}\left(\frac{1}{u}\cdot\frac{\sin u}{\cos u} \right)\;=\;\lim_{u\to0}\left(\frac{\sin u}{u}\cdot\frac{1}{\cos u}\right) \;=\;1\cdot\tfrac{1}{1} \;=\;1
    Thank You Soroban! :]
    Long Time No See

    Im Not A Fan Of Substitution because i never get it right, but you made a good point
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