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Math Help - Trigonometric limit...

  1. #1
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    Trigonometric limit...

    How can I solve this?

    lim t -> 0

    (5t sec4t / tan10t)

    I don't know what to do when the term is secant?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    How can I solve this?

    lim t -> 0

    (5t sec4t / tan10t)

    I don't know what to do when the term is secant?
    \lim_{\varphi\to{0}}\frac{5\varphi\sec(4\varphi)}{  \tan(10\varphi)}

    \tan(10\varphi)\sim{10\varphi}

    Giving us

    \lim_{\varphi\to{0}}\frac{5\varphi\sec(4\varphi)}{  10\varphi}=\lim_{\varphi\to{0}}\frac{5\sec(4\varph  i)}{10}=\frac{1}{2}
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  3. #3
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    I don't understand the second line?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    I don't understand the second line?
    \tan(10\varphi)\sim{10\varphi}

    This means that they are asymptotically equal as \varphi\to{0}

    Or in other \lim_{\varphi\to{0}}\frac{\tan(10\varphi)}{10\varp  hi}=1

    This can be easily proven

    Let \psi=10\varphi

    Now as \varphi\to{0}\Rightarrow\psi\to{0}

    Giving us

    \lim_{\psi\to{0}}\frac{\tan(\psi)}{\psi}=\lim_{\ps  i\to{0}}\frac{\psi+\frac{\psi^3}{3}+...}{\psi}=1

    \therefore\lim_{\varphi\to{0}}\frac{\tan(10\varphi  )}{10\varphi}=1\Rightarrow\frac{\lim_{\varphi\to{0  }}\tan(10\varphi)}{\lim_{\varphi\to{0}}10\varphi}=  1 \Rightarrow\lim_{\varphi\to{0}}\tan(10\varphi)=10\  varphi

    Thus they are interchangable, so I just replaced them

    It's easier because you just know it, you could also use L'hopitals or maclaurin series expansion
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  5. #5
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    Krizalid's Avatar
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    Mathstud28, you may consider what level is theowne currently stuying, he may won't be able to digest your post, so I suggest you to first give a stantard solution and after that, give another one with other somewhat advanced stuff.

    -----

    \underset{t\to 0}{\mathop{\lim }}\,\frac{5t\sec 4t}{\tan 10t}=\frac{1}{2}\underset{t\to 0}{\mathop{\lim }}\,\frac{10t}{\tan 10t}\cdot \sec 4t.

    Now, consider \underset{t\to 0}{\mathop{\lim }}\,\frac{t}{\tan t}=1, which is a little fact that you should know; besides \lim_{t\to0}\sec4t does exist, and its value is 1, hence, the original limit is \frac12.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    How can I solve this?

    lim t -> 0

    (5t sec4t / tan10t)

    I don't know what to do when the term is secant?
    \lim_{x\to{0}}\frac{5x\sec(4x)}{\tan(10x)}=\lim_{x  \to{0}}\sec(4x)\cdot\lim_{x\to{0}}\frac{5x}{\tan(1  0x)}=1\cdot\lim_{x\to{0}}\frac{5x}{\tan(10x)}

    Now there are a coulple of ways of going from here

    Let

    L=\lim_{x\to{0}}\frac{5x}{\tan(10x)}\Rightarrow\fr  ac{1}{L}=\lim_{x\to{0}}\lim_{x\to{0}}\frac{\tan(10  x)}{5x}

    Now extracting a five and seeing that

    \tan(10\cdot{0})=0 and knowing that subtracting zero is legal we have

    \frac{1}{5}\lim_{x\to{0}}\frac{\tan(10x)-\tan(10\cdot{0})}{x-0}

    Now if you notice this looks like the definition of the derivative at a point

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

    so \frac{1}{5}\lim_{x\to{0}}\frac{\tan(10x)-\tan(10\cdot{0})}{x-0}=\frac{1}{5}\bigg(\tan(10x)\bigg)'\bigg|_{x=0} =\frac{1}{5}\bigg(10\sec^2(10x)\bigg)\bigg|_{x=0}=  \frac{1}{5}\cdot{10\sec^2(0)}=2

    So knowing that L is our orginal limit and

    \frac{1}{L}=2\Rightarrow{L=\frac{1}{2}}
    ----------------------------------------------------------------------
    another possibility is L'hopitals

    so we have

    \lim_{x\to{0}}\frac{5x}{\tan(10x)}=\lim_{x\to{0}}\  frac{5}{10\sec^2(10x)}=\frac{5}{10\sec^2(0)}=\frac  {1}{2}
    ----------------------------------------------------------------------
    Another possibility is this

    \lim_{x\to{0}}\frac{5x}{\tan(10x)}=\lim_{x\to{0}}\  frac{\cos(10x)5x}{\sin(10x)}=\lim_{x\to{0}}\cos(10  x)\cdot\lim_{x\to{0}}\frac{5x}{\sin(10x)}=\lim_{x\  to{0}}\frac{5x}{\sin(10x)}

    Now we do this

    Let

    L=\lim_{x\to{0}}\frac{5x}{\sin(10x)}\Rightarrow\fr  ac{2}{L}=\lim_{x\to{0}}\frac{\sin(10x)}{10x}

    Now letting

    \varphi=10x

    we see that as x\to{0}\Rightarrow\varphi\to{0}

    So we would have

    \frac{2}{L}=\lim_{\varphi\to{0}}\frac{\sin(\varphi  )}{\varphi}

    I am sure you know the right hand limit is one

    So we have

    \frac{2}{L}=1\Rightarrow{L=\frac{1}{2}}
    ------------------------------------------------------------
    Whew...there are a couple more but my fingers hurt

    I did these because Krizalid is right and I felt bad so I decided to show you a couple of ways that will be at your level. Hope this helps, sorry about the earlier confusion
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  7. #7
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    I don't really understand what you did in the final post (apart from the l'hopitals rule, which we're not allowed to use). But thanks for the help.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    I don't really understand what you did in the final post (apart from the l'hopitals rule, which we're not allowed to use). But thanks for the help.
    ooooooooooooook then no problem
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