Results 1 to 6 of 6

Math Help - Does the limit as x-> inifinity of cosx exist?

  1. #1
    Member
    Joined
    Jul 2010
    Posts
    86

    Does the limit as x-> inifinity of cosx exist?

    lim x-> infinity (e^(-x)cosx) ?

    First of all does the limit as x-> inifinity of cosx even exist?
    I thought it would = inifinity but then again i feel like it DNE as well.

    Ive atempted this problem in 2 ways, but i first would like to see if cosx as x approaches infinity even exists
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,042
    Thanks
    885
    Quote Originally Posted by kensington View Post
    lim x-> infinity (e^(-x)cosx) ?

    First of all does the limit as x-> inifinity of cosx even exist?
    I thought it would = inifinity but then again i feel like it DNE as well.

    Ive atempted this problem in 2 ways, but i first would like to see if cosx as x approaches infinity even exists

    \displaystyle \lim_{x \to \infty} e^{-x} \cos{x} = 0

    the fact that \displaystyle \lim_{x \to \infty} \cos{x} does not exist is inconsequential in this case ... the value of \cos{x} stays relatively fixed between -1 and 1 ... e^{-x} is the dominant factor here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by kensington View Post
    lim x-> infinity (e^(-x)cosx) ?

    First of all does the limit as x-> inifinity of cosx even exist?
    I thought it would = inifinity but then again i feel like it DNE as well.

    Ive atempted this problem in 2 ways, but i first would like to see if cosx as x approaches infinity even exists
    The trick here is to recall that \forall x\in\mathbb{R},\,-1\leq \cos x\leq 1. This is important because as x\to\infty, \cos x is going to remain finite (but we don't necessarily know the limit value).

    So the only term that will really effect the limit will be the e^{-x} term. So we should see that \lim\limits_{x\to\infty}e^{-x}\cos x\rightarrow 0\cdot (\text{finite number}) = 0.

    Does this make sense?



    EDIT: skeeter beat me to it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2010
    Posts
    86
    Quote Originally Posted by Chris L T521 View Post
    The trick here is to recall that \forall x\in\mathbb{R},\,-1\leq \cos x\leq 1. This is important because as x\to\infty, \cos x is going to remain finite (but we don't necessarily know the limit value).

    So the only term that will really effect the limit will be the e^{-x} term. So we should see that \lim\limits_{x\to\infty}e^{-x}\cos x\rightarrow 0\cdot (\text{finite number}) = 0.

    Does this make sense?
    Ohh because i was simply trying to figure out if we use L'hospitals rule? or you can even rewrite is as cosx/e^x but i was always stuck with the cosx and then e^x would approach infinity in that case.
    So you mierly just "ignored" the cos and did the limit to e^-x which is 0?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,042
    Thanks
    885
    Quote Originally Posted by kensington View Post
    Ohh because i was simply trying to figure out if we use L'hospitals rule? or you can even rewrite is as cosx/e^x but i was always stuck with the cosx and then e^x would approach infinity in that case.
    So you mierly just "ignored" the cos and did the limit to e^-x which is 0?
    I wouldn't call it "ignore" ... it would be easier to see the idea if you wrote the limit as

    \displaystyle \lim_{x \to \infty} \frac{\cos{x}}{e^x}

    as x \to \infty , \cos{x} oscillates between -1 and 1 (i.e. stays relatively fixed) and e^x gets large w/o bound ...

    \displaystyle \frac{value \, between \, -1 \, and \, 1}{very \, large \, value} \to 0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jul 2010
    Posts
    86
    Quote Originally Posted by skeeter View Post
    I wouldn't call it "ignore" ... it would be easier to see the idea if you wrote the limit as

    \displaystyle \lim_{x \to \infty} \frac{\cos{x}}{e^x}

    as x \to \infty , \cos{x} oscillates between -1 and 1 (i.e. stays relatively fixed) and e^x gets large w/o bound ...

    \displaystyle \frac{value \, between \, -1 \, and \, 1}{very \, large \, value} \to 0

    Got it thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Limit of ln(cosx)/x^2
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 27th 2011, 03:52 PM
  2. Limit = 0, or does it exist?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 24th 2010, 06:48 AM
  3. Verify that √((1-cosx)/(1+cosx)) = (1-cosx)/|sinx|
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 10th 2008, 09:39 PM
  4. Limit does not exist
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 6th 2008, 01:50 PM
  5. Does a limit exist?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 6th 2008, 11:31 AM

/mathhelpforum @mathhelpforum