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Math Help - Indeterminate forms - L'Hospitals Rule

  1. #1
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    Indeterminate forms - L'Hospitals Rule

    I'm having trouble with a couple homework problems...can anyone help????

    #1. Find the limit as x approaches 0:

    6x-sin(6x)/6x-tan(6x)

    I know I'm supposed to use L'Hospitals rule here, but when I do it, I get 1, which is apparently wrong. I think I might be differentiating wrong....

    #2. Find the limit as x approaches infinity:

    xtan(8/x)

    Please help!!!
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    Moo
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    Hello,

    Quote Originally Posted by Jessica098 View Post
    I'm having trouble with a couple homework problems...can anyone help????

    #1. Find the limit as x approaches 0:

    6x-sin(6x)/6x-tan(6x)

    I know I'm supposed to use L'Hospitals rule here, but when I do it, I get 1, which is apparently wrong. I think I might be differentiating wrong....
    Use it twice, because after one of this rule, it's still undefined

    #2. Find the limit as x approaches infinity:

    xtan(8/x)

    Please help!!!
    Change the variable :

    t=8/x --> x=8/t

    Hence \lim_{x \to \infty} x \tan \frac 8x=\lim_{t \to 0^+} 8 \frac{\tan x}{x}=8 \cdot \lim_{t \to 0^+} \frac{\tan x -\overbrace{\tan 0}^{=0}}{x-0}

    This is the derivative number of the function tan at x=0. The derivative of \tan x is \frac{1}{\cos^2 x}

    Therefore, we have :

    \lim_{x \to \infty} x \tan \frac 8x=8 (\tan x)'_{x=0}=\frac{8}{\cos^2 0}=8
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  3. #3
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    Quote Originally Posted by Jessica098 View Post
    I'm having trouble with a couple homework problems...can anyone help????

    #1. Find the limit as x approaches 0:

    6x-sin(6x)/6x-tan(6x)

    I know I'm supposed to use L'Hospitals rule here, but when I do it, I get 1, which is apparently wrong. I think I might be differentiating wrong....
    Let us find out!

    f(x) = 6x - \sin(6x)

    f'(x) = 6 - \cos(6x)*6

    g(x) = 6x - \tan(6x)

    g'(x) = 6 - \sec^2(6x)*6

    \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \frac{0}{0}

    That is another indeterminate form, so we need to apply L'Hopital's rule once again..

    f''(x) = 36\sin(6x)

    g''(x) = 72\sec^2(6x)\tan(6x) \Rightarrow \frac{72\sin(6x)}{\cos^3(6x)}

    \frac{f''(x)}{g''(x)} = \frac{36\sin(6x)}{\frac{72\sin(6x)}{\cos^3(6x)}} \Rightarrow \frac{\cos^3(6x)}{2}

    \lim_{x \rightarrow 0} \frac{f''(x)}{g''(x)} = \frac{1}{2}
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  4. #4
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    Quote Originally Posted by Jessica098 View Post
    6x-sin(6x)/6x-tan(6x)
    You should not be this far along and still have notation this sloppy.

    If you mean \frac{6x - sin(x)}{6x - tan(x)}, then you should write that, remembering all that time you spent on the Order of Operations back before Algebra I.

    Maybe add some grouping symbols? [6x-sin(6x)]/[6x-tan(6x)]

    Note: Really, I am not mocking or being demeaning, but I am trying to get your attention and the attention of anyone reading this post. It IS important to write clearly and to write what you mean. This removes the need for guessing what you mean for those who would like to help. Still friends?
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by colby2152 View Post
    Let us find out!

    f(x) = 6x - \sin(6x)

    f'(x) = 6 - \cos(6x)*6

    g(x) = 6x - \tan(6x)

    g'(x) = 6 - \sec^2(6x)*6

    \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \frac{0}{0}

    That is another indeterminate form, so we need to apply L'Hopital's rule once again..

    f''(x) = 36\sin(6x)

    g''(x) = 72\sec^2(6x)\tan(6x) \Rightarrow \frac{72\sin(6x)}{\cos^3(6x)}

    \frac{f''(x)}{g''(x)} = \frac{36\sin(6x)}{\frac{72\sin(6x)}{\cos^3(6x)}} \Rightarrow \frac{\cos^3(6x)}{2}

    \lim_{x \rightarrow 0} \frac{f''(x)}{g''(x)} = \frac{1}{2}
    If this is indeed

    \lim_{x\to{0}}\frac{6x-\sin(x)}{6x-\tan(x)}

    The answer is 1...

    As x\to{0} we see that \sin(x)\sim{x} and \tan(x)\sim{x}

    So \lim_{x\to{0}}\frac{6x-x}{6x-x}=1

    Could someone second me or am I making a mistake?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by colby2152 View Post
    \lim_{x \rightarrow 0} \frac{f''(x)}{g''(x)} = \frac{1}{2}
    Au contrare!
    f''(x) = -36~sin(6x)
    so the limit is -1/2.

    Quote Originally Posted by Mathstud28 View Post

    So \lim_{x\to{0}}\frac{6x-x}{6x-x}=1
    Please read the question more carefully. The problem is
    \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)}

    so your method gives
    \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)} = \lim_{x \to 0} \frac{6x - (6x - 36x^3)}{(6x - (6x + 72x^3)}

    = \frac{36}{-72} = -\frac{1}{2}

    -Dan
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by topsquark View Post
    Au contrare!
    f''(x) = -36~sin(6x)
    so the limit is -1/2.


    Please read the question more carefully. The problem is
    \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)}

    so your method gives
    \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)} = \lim_{x \to 0} \frac{6x - (6x - 36x^3)}{(6x - (6x + 72x^3)}

    = \frac{36}{-72} = -\frac{1}{2}

    -Dan
    Quote Originally Posted by TKHunny View Post
    You should not be this far along and still have notation this sloppy.

    If you mean \frac{6x - sin(x)}{6x - tan(x)}, then you should write that, remembering all that time you spent on the Order of Operations back before Algebra I.

    Maybe add some grouping symbols? [6x-sin(6x)]/[6x-tan(6x)]

    Note: Really, I am not mocking or being demeaning, but I am trying to get your attention and the attention of anyone reading this post. It IS important to write clearly and to write what you mean. This removes the need for guessing what you mean for those who would like to help. Still friends?
    Sorry this post above threw me off
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    If this is indeed

    \lim_{x\to{0}}\frac{6x-\sin(x)}{6x-\tan(x)}

    The answer is 1...

    As x\to{0} we see that \sin(x)\sim{x} and \tan(x)\sim{x}

    So \lim_{x\to{0}}\frac{6x-x}{6x-x}=1

    Could someone second me or am I making a mistake?
    If we linearly approximate sinx at x=0, we get:

    L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=sin(0)+cos(0)(x-0)=\color{red}\boxed{x}

    If we linearly approximate tanx at x=0, we get:

    L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=tan(0)+sec^2(0)(x-0)=\color{red}\boxed{x}

    Then we have:

    \lim_{x \to{0}}\frac{6x-x}{6x-x}=1.

    It looks good to me, unless I too am making a mistake!
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    If we linearly approximate sinx at x=0, we get:

    L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=sin(0)+cos(0)(x-0)=\color{red}\boxed{x}

    If we linearly approximate tanx at x=0, we get:

    L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=tan(0)+sec^2(0)(x-0)=\color{red}\boxed{x}

    Then we have:

    \lim_{x \to{0}}\frac{6x-x}{6x-x}=1.

    It looks good to me, unless I too am making a mistake!

    Haha why do it that way?

    \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra  c{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{x}=\lim_{x\to{0}}1-\frac{x^2}{3!}+\frac{x^4}{5!}-...=1

    \therefore\text{As}\text{ }x\to{0},\sin(x)\sim{x}

    Why do a linear approximation when you can use series?

    Its much simpler....although I would say that your method is unique
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  10. #10
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Haha why do it that way?

    \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra  c{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{x}=\lim_{x\to{0}}1-\frac{x^2}{3!}+\frac{x^4}{5!}-...=1

    \therefore\text{As}\text{ }x\to{0},\sin(x)\sim{x}

    Why do a linear approximation when you can use series?

    Its much simpler....although I would say that your method is unique
    Well, Series is not my specialty...I thought this was simpler my way...I think my way looks nicer

    I agree that your's works as well...but I like mine better.
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  11. #11
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    I'm pretty sure that Jessica is not ready to diggest post #9, since she's covering differentiation.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    I'm pretty sure that Jessica is not ready to diggest post #9, since she's covering differentiation.
    ....Thats what I would think but someone informed me on here (one of the bigwigs) that most people learn series in conjunction with power series? Dont ask me, that seems absurd.

    But alternatively

    Utilizing L'hopitals

    \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra  c{\cos(x)}{1}=1

    and \lim_{x\to{0}}\frac{\tan(x)}{x}=\lim_{x\to{0}}\fra  c{\sec^2(x)}{1}=1

    These both because the original gave an indeterminate form

    also

    \lim_{x\to{0}}\frac{sin(x)}{x}=\lim_{x\to{0}}\frac  {\sin(x)-\sin(0)}{x-0}

    and by definition of a derivative at a point

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

    we see that \lim_{x\to{c}}\frac{\sin(x)-\sin(0)}{x-0}=\bigg(\sin(x)\bigg)'\bigg|_{x=0}=\cos(0)=1


    Likewise

    \lim_{x\to{0}}\frac{\tan(x)}{x}=\lim_{x\to{0}}\fra  c{\tan(x)-\tan(0)}{x-0}

    and by the same logic as the previous

    \lim_{x\to{0}}\frac{\tan(x)-\tan(0)}{x-0}=\bigg(\tan(x)\bigg)'\bigg|_{x=0}\sec^2(0)=1


    Both of those methods validate that

    \text{As}\text{ }x\to{0}\text{ }\sin(x)\sim{x}

    as well as \text{As}\text{ }x\to{0}\text{ }\tan(x)\sim{x}
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  13. #13
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    Show off...


    @ topsquark : au contraIre ^^

    @ guys : no... nothing
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