Results 1 to 10 of 10

Math Help - death by limits!!!

  1. #1
    Newbie
    Joined
    Sep 2006
    From
    Chile
    Posts
    10

    death by limits!!!

    hello... i just started university and in my math class we are seing limits and since here it said limits well i decided to post it here....

    well ive been having lots of trouble with trigonometry stuff and now they go and mix one problem with limits

    Im supposed to find:

    lim {cos^2(x/2)}/{cos(x)-sen(3x/2)}
    x->pi

    thanks for the help...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    What does "sen" mean? secant?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2006
    From
    Chile
    Posts
    10

    death by limits!!!

    amm yeah... sen => sin ...... its in spanish and forgot to change it...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by sacwchiri View Post
    hello... i just started university and in my math class we are seing limits and since here it said limits well i decided to post it here....

    well ive been having lots of trouble with trigonometry stuff and now they go and mix one problem with limits

    Im supposed to find:

    lim {cos^2(x/2)}/{cos(x)-sen(3x/2)}
    x->pi

    thanks for the help...
    We could use L'Hopital's rule here (i don't think you would have learnt it at this time since this question was asked apparently at the begining or before your calculus class, but i can't see a way to factor things at the moment).

    To refresh your memory:

    whenever we have the cases:

    \lim_{x \to a} \frac {f(x)}{g(x)} = \frac {0}{0} \mbox { or } \frac {\infty}{\infty}

    we can apply L'Hopital's rule (several times, if needed) to find the limit.


    L'Hopital's Rule:
    \lim_{x \to a} \frac {f(x)}{g(x)} = \lim_{x \to a} \frac {f'(x)}{g'(x)}




    Now, on to your question:

    Let L = \lim_{x \to \pi} \frac {\cos^2 \left( \frac {x}{2} \right)}{\cos x - \sin \left( \frac {3x}{2} \right)} ..........we get \frac {0}{0} if we plug in x = \pi

    Apply L'Hopital's, we get:

    L = \lim_{x \to \pi} \frac { \cos \left( \frac {x}{2} \right) \sin \left( \frac {x}{2} \right) }{ \sin x + \frac {3}{2} \cos \left( \frac {3x}{2} \right) } .........again, we get \frac {0}{0} we we plug in x = \pi

    Apply L'Hopital's again, we get:

    L = \lim_{x \to \pi} \frac { \frac {1}{2} \left( \cos^2 \left( \frac {x}{2} \right) - \sin^2 \left( \frac {x}{2} \right) \right)}{ \cos x - \frac {9}{4} \sin \left( \frac {3x}{2} \right)} ........no problems here when we plug in x = \pi

    \Rightarrow L = \frac {- \frac {1}{2}}{ \frac {5}{4}} = - \frac {2}{5}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I do not think L'Hopital's rule can be used here. See what happens if you substitute, or try to substitute. You get 0.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Actually in order to solve this problem you must use L'Hospital's rule, and Jhevon got the correct answer.




    -qbkr21
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by qbkr21 View Post
    Actually in order to solve this problem you must use L'Hospital's rule, and Jhevon got the correct answer.




    -qbkr21
    I don't like the word "must." Just because we can't see another way to do it doesn't mean one doesn't exist!

    And, after all, you can do it numerically and approach \pi from both sides and at least make a case for -2/5 without doing anything fancy. Perhaps that was all that's required here for this class.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by topsquark View Post
    I don't like the word "must." Just because we can't see another way to do it doesn't mean one doesn't exist!

    And, after all, you can do it numerically and approach \pi from both sides and at least make a case for -2/5 without doing anything fancy. Perhaps that was all that's required here for this class.

    -Dan
    yes, i forgot about that method...i hate it though, it's too calculation intensive and it gets annoying if you don't have a calculator that you can program the formula into
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Dan that is a good point however I would never go about doing a limits problem that way unless one was shooting up to - \infty or \infty. It would be one of those limit approaches constant problems where you plug the constant in and you dealing with fraction divided by zero.


    -qbkr21
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qbkr21 View Post
    Dan that is a good point however I would never go about doing a limits problem that way unless one was shooting up to - \infty or \infty. It would be one of those limit approaches constant problems where you plug the constant in and you dealing with fraction divided by zero.


    -qbkr21
    the method Dan was talking about can be used even if it is not an infinite limit. finding the limit by the numerical method means you plug in numbers increasing close to the number that causes the function to be undefined from both directions.

    say you wanted to evaluate a limit as x approaches 2, but when x is 2 the function is undefined. you could plug in numbers close to 2 from the right and the left, and if the result seems to settle at a number as you get close to 2, then that is the limit, if it doesn't, then the limit does not exist. so you would try numbers like:

    1.9
    1.99
    1.999
    1.9999
    1.99999
    (skip 2)
    2.01
    2.001
    2.0001
    2.00001

    if the result settles at a number, as we get closer to 2, then that's the limit

    so for this problem, we would try, say:

    3.1412
    3.1413
    3.1414
    (skip 3.1415 ~= pi)
    3.1416
    3.1417
    3.1418
    3.15

    if we do this method, we would see the result settling at -0.4 as we get close to pi. so that would be our limit. but as i said before, pluging in all those values into the formula given and calculating each result to see if they settle somewhere would be a pain. factoring algebraically would be much more fun and easier, but i couldn't see a way to do that, so i did L'Hopital's
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Birth and Death Process
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 28th 2010, 06:13 PM
  2. Birth and Death Process
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 27th 2010, 08:56 PM
  3. Death Rates
    Posted in the Business Math Forum
    Replies: 0
    Last Post: September 27th 2009, 08:40 AM
  4. Simple birth-death process
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: June 14th 2009, 06:14 AM
  5. Pure death process
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 10th 2009, 09:33 AM

Search Tags


/mathhelpforum @mathhelpforum