Results 1 to 5 of 5

Math Help - Limit Comprehension Question.

  1. #1
    Junior Member
    Joined
    Nov 2012
    From
    Michigan
    Posts
    27
    Thanks
    1

    Limit Comprehension Question.

    The limit problem is as follows: \lim_{x\to\0}\frac{11x}{tan21x}.
    Some thoughts I had about solving this were to express the inverse of tangent as a fraction of cosine divided by sine: \frac{cos11x}{sin21x}. By doing this I thought that I could then use the \lim_{x\to\0}\frac{sin(x)}{x}=1 theorem(?) to multiply the numerator by 21 yielding \frac{21x}{sin21x} which equals one. But I don't really know what to do with cosine and this is where I am stuck. The answer in the book is \frac{11}{21}. If someone could help point me in the right direction I would really appreciate it.


    (P.S., sorry if this is in the wrong forum)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member Barioth's Avatar
    Joined
    Jan 2013
    From
    Canada
    Posts
    54
    Thanks
    16

    Re: Limit Comprehension Question.

    You solved it already.

    since tan(21x) = \frac{\sin(21x)}{\cos(21x)} you end up with

    \lim_{x\to\0}{\frac{11x*cos21x}{sin21x}}
     = 11* \lim_{x\to\0}{\frac{ 21x *\cos(21x)}{21*\sin(21x)}}
     = 11* \lim_{x\to\0}{\frac{1* \cos(21x)}{21}} .. .. .. .. Since you said \lim_{x\to\0} {\frac{21x}{sin(21x} = 1
    = 11*\frac{1}{21} * \lim_{x\to\0}{\cos(21x)} .. .. .. .. since  \cos(21*0) = 1
    =\frac{11}{21}

    Have a good day
    Last edited by Barioth; January 13th 2013 at 08:15 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,908
    Thanks
    766

    Re: Limit Comprehension Question.

    Hello, KhanDisciple!

    You have the right idea.
    We just need some Olympic-level gymnastics.


    \lim_{x\to0}\frac{11x}{\tan(21x)}

    We have: . \frac{11x}{\tan(21x)} \;=\; \frac{11x}{\frac{\sin(21x)}{\cos(21x)}} \;=\;\frac{11x\cos(21x)}{\sin(21x)}

    Multiply by \frac{21}{21}:\;\;\frac{21}{21}\cdot\frac{11x\cos(  21x)}{\sin(21x)} \;=\;\frac{11}{21}\cdot\frac{21x}{\sin(21x)}\cdot \cos(21x)


    Therefore: . \lim_{x\to0}\left[\frac{11}{21}\cdot\frac{21x}{\sin(21x)}\cdot\cos(2  1x)\right] \;=\;\frac{11}{21}\cdot 1 \cdot 1 \;=\;\frac{11}{21}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602

    Re: Limit Comprehension Question.

    Quote Originally Posted by KhanDisciple View Post
    The limit problem is as follows: \lim_{x\to\0}\frac{11x}{tan21x}.
    Some thoughts I had about solving this were to express the inverse of tangent as a fraction of cosine divided by sine: \frac{cos11x}{sin21x}. By doing this I thought that I could then use the \lim_{x\to\0}\frac{sin(x)}{x}=1 theorem(?) to multiply the numerator by 21 yielding \frac{21x}{sin21x} which equals one. But I don't really know what to do with cosine and this is where I am stuck. The answer in the book is \frac{11}{21}. If someone could help point me in the right direction I would really appreciate it.


    (P.S., sorry if this is in the wrong forum)
    If you didn't want to perform the Olympic Level gymnastics, L'Hospital's Rule simplifies things greatly since this goes to \displaystyle \begin{align*} \frac{0}{0} \end{align*}.

    \displaystyle \begin{align*} \lim_{x \to 0}\frac{11x}{\tan{\left( 21x \right)}} &= \lim_{x \to 0}\frac{\frac{d}{dx} \left( 11x \right)}{\frac{d}{dx} \left[ \tan{\left( 21x \right)} \right]} \\ &= \lim_{x \to 0} \frac{11}{21\sec^2{\left( 21x \right)}} \\ &= \frac{11}{21} \end{align*}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2012
    From
    Michigan
    Posts
    27
    Thanks
    1

    Re: Limit Comprehension Question.

    Thanks a lot guys, you are the best!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Joint Distribution Comprehension Issues
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: June 6th 2012, 11:35 PM
  2. Set Term Comprehension
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 19th 2011, 01:09 PM
  3. comprehension -/sinx-cosx/=√1-sin2x
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: July 17th 2011, 05:08 AM
  4. [SOLVED] Correct use of Axiom Schema of Comprehension
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 20th 2011, 03:51 PM
  5. Bounded Comprehension Principle
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: January 19th 2011, 09:39 PM

Search Tags


/mathhelpforum @mathhelpforum