Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calc 101 Limit Question

  1. Sep 9, 2004 #1
    Lim 1 - cos(x)/x^2

    My attempt:

    lim [1 - cos (x)/ x][lim 1/x] = (1)(1/0) :cry:

    Please help me with some basic examples.
  2. jcsd
  3. Sep 9, 2004 #2
    lim [1 - cos (x)/ x]=1?
    How ??
    While solving limit Problems, you make use of certain solutions
    [tex]\lim_{x\rightarrow 0} \frac{Sin(x)}{x}=1[/tex]

    [tex]\lim_{x\rightarrow 0} \frac{e^x-1}{x}=1[/tex]

    eg:[tex]\lim_{x\rightarrow 0} \frac{Tan(x)}{x}[/tex]

    =>[tex]\lim_{x\rightarrow 0} \frac{\frac{Sin(x)}{cosx}}{x}[/tex]

    ==>[tex]\lim_{x\rightarrow 0} \frac{Sin(x)}{x} \lim_{x\rightarrow 0} \frac{1}{Cosx}[/tex]
    end example==================
    In this case,try to convert this into a known limit
    Last edited: Sep 9, 2004
  4. Sep 9, 2004 #3
    use le'hopitals(spelling?) rule
    Last edited: Sep 9, 2004
  5. Sep 9, 2004 #4


    User Avatar

    Staff: Mentor

    Hmm, I got 1/2. I think you left the x^2 out of the original denominator:

    lim x-->0 of ( 1-cos(x) )/x^2

    = lim x-->0 ( sin(x) )/2x

    = lim x-->0 ( cos(x) )/2 = 1/2

    BTW, here's a good page on L'Hopital's Rule:


    PS -- I went back and used a calculator to plug in small numbers for x, and it looks more like the limit of the original function of x goes to zero. I wonder if I did the first differentiation wrong, or the calculator is fooling me. What do other folks get?
    Last edited: Sep 9, 2004
  6. Sep 9, 2004 #5

    I guess this can be solved in eitherof following 2 ways.:

    1. Using formula of cosx=1-(x^2)/2!+(x^4)/4!...
    so that
    so that
    Lt x->0 of (1-cosx)/x^2 becomes :
    =Lt x->0 of 1/2[1-(x^2)/4!....]

    2. other method could be usin L'Hospital's rule using differentiation
    this too yields answer as 1/2

    take care :smile:
  7. Sep 9, 2004 #6
    Your calculator was probably using degrees instead of radians.
  8. Sep 9, 2004 #7


    User Avatar

    Staff: Mentor

    >Your calculator was probably using degrees instead of radians.

    Perfect! Thanks, Muzza. I thought I was losing my mind!

    I was going to suggest to Halcyon99 that a good way to check the results you get from L'Hospital's Rule is to just plug some small numbers into the original fractional expression using a calculator. Now I know to add the caveat that the calculator needs to be in Radian mode. I'd spaced the part about when series expansions or other expressions are shown sharing a variable between trig functions and algebraic functions, the trig function arguments are in radians.

    Thanks again, -Mike-
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook