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!

Limits homework

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    IF 5X<=f(X)<=X^3+2
    FIND LIM F(X)IS X=2
     
  2. jcsd
  3. Sep 12, 2009 #2
    Re: Limits

    I'm sorry, what are you asking? Is it this:

    If [itex]5x\le f(x)\le x^3+2[/itex] then find [itex]\lim_{x\rightarrow2}f(x)[/itex] ?
     
  4. Sep 12, 2009 #3
    Re: Limits

    yes that is correct
     
  5. Sep 12, 2009 #4
    Re: Limits

    What do you know about the limit of a function that is being "Sandwiched" or "Squeezed" between two other functions? hint...hint...:wink:
     
  6. Sep 12, 2009 #5
    Re: Limits

    so I have 10<=2<=10 and the equations equals 10
     
  7. Sep 12, 2009 #6
    Re: Limits

    Yes. :smile:
     
  8. Sep 12, 2009 #7
    Re: Limits

    I think I understand limits but this last question says that lim x=0
    evaluate limx^6 cos 7/x
    I came up with 0 or would it be limit does not exist.
     
  9. Sep 12, 2009 #8
    Re: Limits

    [tex]\lim_{x\rightarrow0}x^6\cos(\frac{7}{x})[/tex] ?
     
  10. Sep 12, 2009 #9
    Re: Limits

    yes that is correct
     
  11. Sep 12, 2009 #10
    Re: Limits

    Hmmm... I am not sure. Perhaps someone else can chime in here. I don't remember all of the 'tricks' to evaluating limits.

    If you can find a way to rewrite the function such that the 'x' in the denominator can be cancelled, then the limit exists.

    I would play around with the trig identities maybe?
     
  12. Sep 13, 2009 #11
    Re: Limits

    The limit should be 0, since x gets arbitrarily small. Although there is oscillation from the cos, the function is bounded. You should try proving this rigorously.
     
  13. Sep 13, 2009 #12

    VietDao29

    User Avatar
    Homework Helper

    Re: Limits

    No.. you don't need any trig identity here. When dealing with limits of this kind, i.e involving the product of one part tending to 0, and the other part oscillating. Then, one should think right about the range of the oscillating part, and how "big" can it get.

    Hint hint.. :wink:

     
  14. Sep 13, 2009 #13
    Re: Limits

    Is one of the 'factors' oscillating though? It looks undefined. But, like I said, it's been awhile :smile:

    [tex]
    \lim_{x\rightarrow0}x^6\cos(\frac{7}{x})
    [/tex]
     
  15. Sep 13, 2009 #14

    VietDao29

    User Avatar
    Homework Helper

    Re: Limits

    The function is undefined at x = 0. However the limit does exist when x tends to 0.

    Another example is:

    [tex]\frac{x ^ 2}{x}[/tex] is undefined at x = 0. But:

    [tex]\lim_{x \rightarrow 0} \frac{x ^ 2}{x} = \lim_{x \rightarrow 0} x = 0[/tex].

    ------------------

    Or:

    [tex]\frac{\sin(x)}{x}[/tex] is undefined at x = 0. But:

    [tex]\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1[/tex].

    Because, when taking limit as x tends to 0. We just consider the values for x near 0 (0 excluded). So, when the function is undefined at x = 0, the limit may still exist. :)

    Well, if you are still unsure about how to do it. I'll give you a hint then:

    1. Find the range for cos(x)
    2. Then, try to apply the "Sandwich Theorem" like this:
      [tex]... \leq x ^ 6 \cos\left(\frac{7}{x} \right) \leq ...[/tex]
     
  16. Sep 13, 2009 #15
    Re: Limits

    Ah. I see, I am actually just reviewing limits fir the first time in years. I have past the section on how to deal with rational functions with zero denominators if the zero in the denominator can be 'eliminated.'

    But, have yet to get to the part where they cannot.

    Thanks for the preview :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limits homework
  1. Homework help limits? (Replies: 3)

  2. Limit homework issue (Replies: 3)

Loading...