1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit question

Tags:
  1. Nov 7, 2016 #1
    1. The problem statement, all variables and given/known data
    can you cancel the x and the |x|


    lim x→0 ( x(1-(cos(x))/|x| )

    2. Relevant equations



    3. The attempt at a solution
    At first i thought the limit would just be undefined as x approaches 0 but the answer to the problem is actually 0, so can you just cancel the x with the magnitude of x (|x|) on the denominator??? thanks
     
  2. jcsd
  3. Nov 7, 2016 #2

    Mark44

    Staff: Mentor

    No, you can't just cancel x with |x|. If x > 0, ##\frac x {|x|} = 1##, but if x < 0, ##\frac x {|x|} = -1##. You need to try something else.
     
  4. Nov 7, 2016 #3
    how about if you expand the top to x-xcos(x) for the first step but then youre still left with x-xcos(x)/|x|, is that the correct first step? how would i continue from here, not really sure because you cant have that denominator there since as x approaches 0 the limit is going to be undefined right??
     
  5. Nov 7, 2016 #4

    Mark44

    Staff: Mentor

    No, that won't work -- you still have something of the form ##\frac 0 0 ##.

    Hint: ##\frac x {|x|}## is the same as ##\frac{|x|} x##. Then write the expression as the product of two other expressions. I can't really say much more without giving the solution away.
     
  6. Nov 7, 2016 #5
    im a little confused but, youre saying i can just swap the x in the numerator with the |x| in the denominator, and get |x|-xcos(x)/x or |x|/x⋅(1-cos(x)) ??
     
  7. Nov 7, 2016 #6
    somewhere youre going to have to cancel an x correct ? to get it out of 0/0 form
     
  8. Nov 7, 2016 #7

    Mark44

    Staff: Mentor

    Neither one is equal to the expression you started with. The first one you wrote just above is ##|x| - \frac{x\cos(x)}x##. The second is ##\frac{|x|}x - (1 - \cos(x))##.
     
  9. Nov 7, 2016 #8
    can you explain how i would solve this then please
     
  10. Nov 7, 2016 #9

    Mark44

    Staff: Mentor

    There are a couple of well-known trig limits, one of which can be used in this problem.
     
  11. Nov 7, 2016 #10
    are you meaning as x approaches 0 for cos(x) then cos(x)=1
     
  12. Nov 8, 2016 #11

    Mark44

    Staff: Mentor

    No, that's not the one. It's slightly more complicated.
     
  13. Nov 8, 2016 #12

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You can use l'Hospital's rule to evaluate ##\lim_{x \to 0} \frac{1 - \cos x}{x}##. That eliminates the 0/0 problem, and that is why l'Hospital invented the method. And that is why every calculus book ever written has a section on it.
     
  14. Nov 8, 2016 #13
    okay that was probably the most helpful thing post yet, so then lim x → 0 ( x(1-(cos(x))/|x| ) the derivative of the bottom is going to be 1 and the derivative of the top will be found using chain rule correct? so if we take (1-cos(x)) = u then dy/dx of the numerator will be dy/dx (x) ⋅ dy/dx (1-cos(x)) and we know derivative of x will be 1 so its 1⋅(0-(-sin(x))) therefore, dy/dx of numerator is 1⋅sin(x) and sin(0)=0 and 1⋅0=0 so the answer is 0/1 which is 0?
     
    Last edited by a moderator: Nov 8, 2016
  15. Nov 8, 2016 #14
    could have potentially messed up my derivative skillz there, will it in fact be, dy/dx of the numerator = (1-cos(x))⋅(sin(x)) or was i correct initially ?
     
  16. Nov 8, 2016 #15

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The absolute value sign messes up things a bit (being non-differentiable at the origin); that is why one should use the fact that ##x/|x| = |x|/x##, to move the absolute value sign up into the numerator. Then one can look separately at ##\lim_{x \to 0} |x|## and ##\lim_{x \to 0} [1-\cos x]/x##; now both the numerator and denominator are differentiable in the second limiting ratio.

    Alternatively, we could take the two cases ##x > 0## and ##x < 0##, and look separately at limits from the left and the right.
     
  17. Nov 8, 2016 #16

    Mark44

    Staff: Mentor

    Most, or possibly all, calculus books discuss that particular limit, as well as ##\lim_{x \to 0}\frac{\sin(x)}x## and ##\lim_{x \to 0}\frac{\tan(x)}x##. If you know those limits, there's no need to use L'Hopital's Rule.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit question
  1. Limits question (Replies: 5)

  2. A question on limits (Replies: 11)

  3. A question on limits? (Replies: 4)

  4. Limits Question (Replies: 5)

  5. Limit question (Replies: 2)

Loading...