Results 1 to 10 of 10

Math Help - limit Problem

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    80

    limit Problem

    lim \frac{1+cos{\pi}x}{{tan^2}5{\pi}x} (x \to 1)
    Last edited by gilyos; December 17th 2009 at 07:51 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    See attachment
    Attached Thumbnails Attached Thumbnails limit Problem-limit.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    80
    Thank's man , you think there is another solution without Lupital
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    I would be interested if there were
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    80
    I post this if I find another solution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Quote Originally Posted by gilyos View Post
    Thank's man , you think there is another solution without Lupital
    Sure. How about

     <br />
\frac{1 + \cos \pi x}{\tan^2 5 \pi x}<br />
     <br />
= <br />
(1 + \cos \pi x) \frac{ \cos^2 5 \pi x}{\sin^2 5 \pi x}<br />

     <br />
= <br />
2(1 + \cos \pi x) \frac{ \cos^2 5 \pi x}{1 - \cos 10 \pi x}<br />

     <br />
= <br />
2\cos^2 5 \pi x (1 + \cos \pi x) \cdot \frac{ 1 - \cos \pi x}{1 - \cos \pi x}\cdot \frac{ 1}{1 - \cos 10 \pi x} \cdot \frac{ 1 + \cos 10 \pi x}{1 + \cos 10 \pi x}<br />

     <br />
= <br />
2\cos^2 5 \pi x \cdot \frac{ \sin^2 \pi x}{1 - \cos \pi x}\cdot \frac{ 1 + \cos 10 \pi x}{ \sin^2 10 \pi x}

    so

     <br />
\lim_{x \to 1} <br />
2\cos^2 5 \pi x \cdot \left(\frac{ \sin \pi x}{x}\right)^2 \cdot \left( \frac{x}{\sin^2 10 \pi x}\right)^2\cdot \frac{ 1 + \cos 10 \pi x}{1 - \cos \pi x} = 2(-1)^2 \frac{\pi^2}{100 \pi^2} \cdot \frac{2}{2} = \frac{1}{50}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    \lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{x\to \pi}\frac{1+\cos x}{\tan^2 5x}=\lim_{t\to 0}\frac{1+\cos \left(\pi-t\right)}{\tan^2 5\left(\pi-t\right)}=\lim_{t\to 0}\frac{1-\cos t}{\tan^25t}

    so,

    \lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{t\to 0}\frac{1-\cos t}{t^2}\cdot\frac{25t^2}{\sin^2 5t}\cdot\frac{\cos^2 5t}{25}=\frac{1}{2}\cdot 1 \cdot\frac{1}{25}=\frac{1}{50}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Quote Originally Posted by Abu-Khalil View Post
    \lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{x\to \pi}\frac{1+\cos x}{\tan^2 5x}=\lim_{t\to 0}\frac{1+\cos \left(\pi-t\right)}{\tan^2 5\left(\pi-t\right)}=\lim_{t\to 0}\frac{1-\cos t}{\tan^25t}

    so,

    \lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{t\to 0}\frac{1-\cos t}{t^2}\cdot\frac{25t^2}{\sin^2 5t}\cdot\frac{\cos^2 5t}{25}=\frac{1}{2}\cdot 1 \cdot\frac{1}{25}=\frac{1}{50}.
    The scale and shift to the origin was a nice touch.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Dec 2009
    Posts
    80
    Good one !!!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by gilyos View Post
    lim \frac{1+cos{\pi}x}{{tan^2}5{\pi}x} (x \to 1)
    I would have done the cool manipulation way, but it seems that's already been amply covered.

    Let L denote the above limit, then L=\lim_{x\to0}\frac{1+\cos\left(\pi x-\pi\right)}{\tan^2\left(5\pi x-5\pi\right)}=\frac{1-\cos\left(\pi x\right)}{\tan^2\left(5\pi x\right)}. Let z=\pi x so our limit becomes \lim_{z\to0}\frac{1-\cos(z)}{\tan^2(5z)} and it is well known that \cos(z)\underset{z\to0}{\sim} 1-\frac{z^2}{2} and \tan(z)\underset{z\to0}{\sim} z were it follows that L=\lim_{z\to0}\frac{1-\left(1-\tfrac{z^2}{2}\right)}{25z^2}=\frac{1}{50}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: August 13th 2010, 01:03 AM
  2. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  3. A limit problem
    Posted in the Calculus Forum
    Replies: 8
    Last Post: September 6th 2009, 01:37 AM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  5. limit e problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 25th 2007, 08:04 AM

Search Tags


/mathhelpforum @mathhelpforum