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Math Help - Limit of trig?

  1. #1
    Junior Member Coco87's Avatar
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    Limit of trig?

    Hey,
    This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
    The problem is: \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\  theta}}

    I know the answer is: \frac{1}{2}

    I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

    I get as far as: \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\  theta\cos{\theta}+\sin{\theta}}

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Coco87 View Post
    Hey,
    This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
    The problem is: \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\  theta}}

    I know the answer is: \frac{1}{2}

    I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

    I get as far as: \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\  theta\cos{\theta}+\sin{\theta}}

    Thanks!
    How about using L'Hopital's? Is that allowed?

    Mathstud will give a series solution shortly xD
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Coco87 View Post
    Hey,
    This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
    The problem is: \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\  theta}}

    I know the answer is: \frac{1}{2}

    I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

    I get as far as: \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\  theta\cos{\theta}+\sin{\theta}}

    Thanks!
    \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}

    x+\tan(x)=x+\bigg[x+\frac{1}{3}x^3+\frac{2}{15}x^5+...\bigg]=2x+\frac{1}{3}x^3+\frac{2}{15}x^5\sim{2x}

    \therefore\lim_{x\to{0}}\frac{sin(x)}{x+\tan(x)}=\  lim_{x\to{0}}\frac{x}{2x}=\frac{1}{2}

    Sigh..I am caugh in a stereotype
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  4. #4
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    Quote Originally Posted by Coco87 View Post
    Hey,
    This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
    The problem is: \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\  theta}}

    I know the answer is: \frac{1}{2}

    I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

    I get as far as: \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\  theta\cos{\theta}+\sin{\theta}}

    Thanks!
    \frac{\sin \theta}{\theta + \tan \theta} = \frac{\frac{\sin \theta}{\theta}}{1 + \frac{\tan \theta}{\theta}}.

    Now use the fact, \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = \lim_{\theta \to 0}\frac{\tan \theta}{\theta} = 1.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Coco87 View Post
    Hey,
    This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
    The problem is: \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\  theta}}

    I know the answer is: \frac{1}{2}

    I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

    I get as far as: \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\  theta\cos{\theta}+\sin{\theta}}

    Thanks!
    Alternatively,

    \lim_{x\to{0}}\frac{\sin(x)}{x+\tan(x)}=\lim_{x\to  {0}}\frac{\frac{\sin(x)}{x}}{1+\frac{\tan(x)}{x}}

    Now since it is commonly known that \lim_{x\to{0}}\frac{\sin(x)}{x}=1
    so then

    \lim_{x\to{0}}\frac{tan(x)}{x}=\lim_{x\to{0}}\frac  {\sin(x)}{\cos(x)x}=1\cdot\lim_{x\to{0}}\frac{\sin  (x)}{x}=1

    or you can also do this as rewriting it as

    \lim_{x\to{0}}\frac{\tan(x)-\tan(0)}{x-0} and seeing the defiition of the derivative


    either way we ahve

    \lim_{x\to{0}}\frac{\frac{\sin(x)}{x}}{x+\frac{\ta  n(x)}{x}}=\frac{1}{1+1}=\frac{1}{2}
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  6. #6
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    <br />
  \begin{aligned}\mathop {\lim }\limits_{\theta  \to 0} \frac{{\sin \theta }}<br />
{{\theta  + \tan \theta }} &= \mathop {\lim }\limits_{\theta  \to 0} \left\{ {\frac{{\sin \theta }}<br />
{\theta }\cdot \frac{\theta }<br />
{{\theta +\tan\theta }}} \right\}\\<br />
&= \mathop {\lim }\limits_{\theta  \to 0} \frac{{\sin \theta }}<br />
{\theta } \cdot \mathop {\lim }\limits_{\theta\to 0} \frac{1}<br />
{{1 + \dfrac{{\tan \theta }}<br />
{\theta }}}\\<br />
&= \frac{1}<br />
{2}.\end{aligned}


    This is equal to THP's work. While I was posting he beat me to it.

    (Taylor series solution will make no sense to Coco87.)
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Coco87 View Post
    Hey,
    This is a problem in my Calculus book, but I'm pretty sure it's a trig issue.
    The problem is: \lim_{\theta\to0}\frac{\sin{\theta}}{\theta+\tan{\  theta}}

    I know the answer is: \frac{1}{2}

    I've tried running through a list of Properties, Identities, and Formulas; I still can't seem to obtain the correct answer.

    I get as far as: \lim_{\theta\to0}\frac{\sin{\theta}\cos{\theta}}{\  theta\cos{\theta}+\sin{\theta}}

    Thanks!
    Or you could just use L'Hopital's rule. (Ducks the flying rotten fruit.)

    \lim_{\theta \to 0} \frac{sin(\theta)}{\theta + tan(\theta)}

    = \lim_{ \theta \to 0} \frac{cos(\theta)}{1 + sec^2(\theta)}

    = \frac{1}{1 + 1} = \frac{1}{2}

    -Dan
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    <br />
\begin{aligned}\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}<br />
{{\theta + \tan \theta }} &= \mathop {\lim }\limits_{\theta \to 0} \left\{ {\frac{{\sin \theta }}<br />
{\theta }\cdot \frac{\theta }<br />
{{\theta +\tan\theta }}} \right\}\\<br />
&= \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}<br />
{\theta } \cdot \mathop {\lim }\limits_{\theta\to 0} \frac{1}<br />
{{1 + \dfrac{{\tan \theta }}<br />
{\theta }}}\\<br />
&= \frac{1}<br />
{2}.\end{aligned}


    This is equal to THP's work. While I was posting he beat me to it.

    (Taylor series solution will make no sense to Coco87.)
    Yes, I know, that is why I posted an alternate solution(same as yours actually), but I had to sate Jhevon's need for power series xD
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ...but I had to sate Jhevon's need for power series xD
    Oh come on! you know you wanted to do it, deep down inside!
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  10. #10
    Junior Member Coco87's Avatar
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    Ahh, I see now.

    Thanks!

    I guess I should have caught that...
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    Oh come on! you know you wanted to do it, deep down inside!
    .....no........not at all...
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Or you could just use L'Hopital's rule. (Ducks the flying rotten fruit.)

    \lim_{\theta \to 0} \frac{sin(\theta)}{\theta + tan(\theta)}

    = \lim_{ \theta \to 0} \frac{cos(\theta)}{1 + sec^2(\theta)}

    = \frac{1}{1 + 1} = \frac{1}{2}

    -Dan
    You know what you made me realize topsquark? I think like a physicist! I've never even done physics!
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    You know what you made me realize topsquark? I think like a physicist! I've never even done physics!
    Nooooo!
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Nooooo!
    Yes, it's true Mathstud... i'm a physicist at heart. i know it's hard to believe, but i can't keep lying to myself anymore!
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