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Homework Help: Differentiable statements

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    can someone help me to see whether the following statements choose or false

    a) if lim(X->infinite)f(x) exists and is finite and lim(X->infinite)f'(x)=b then b=0
    i think it is right but i don't know how to prove it

    b)if lim(X->infinite)f(x) exists and is finite then lim(X->infinite)f'(x)=0
    i have no idea guesses right

    c)lim(X->infinite)f'(x)=0 then lim(X->infinite)f(x) exists
    counterexample: f(x)=lnx

    d))lim(X->infinite)f'(x)=0 then lim(X->infinite)f(x)/x=0
    no idea as well
     
  2. jcsd
  3. Feb 20, 2012 #2

    jbunniii

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    For (a), suppose b > 0. Then f'(x) > b/2 for all sufficiently large x. What does this imply about f(x) for sufficiently large x?

    For (b), think about a function that has jump discontinuities at arbitrarily large values of x. Then f'(x) will not exist at the jumps, so lim(X->infinite)f'(x) won't exist, either. Can you come up with an example like this that still satisfies "lim(X->infinite)f(x) exists and is finite"?

    For (c), your counterexample is fine.

    For (d), given epsilon > 0, you have |f'(x)| < epsilon for all sufficiently large x. Pick a specific such x, call it x0. Then for all x beyond that point, f(x) must be bounded between two lines, of slope +/- epsilon. Work out the equations for those lines, and see what this implies about f(x)/x for x > x0.
     
  4. Feb 20, 2012 #3
    sorry ,for a) i still can understand how to prove it , can u give me more details then cfor b) my example is y=[x] the integer part of x
    and for d) i think that for all sufficiently large x , take x0, for all x beyond that point then then
    -epsilon<f'(x)<+epsilon , then -epsilonx<f(x)>epsilonx then ,f(x)/x for all x0 bounded by epsilon that is 0 does it make sense?
     
  5. Feb 20, 2012 #4

    jbunniii

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    Your counterexample for (b) doesn't work, because [x] goes to infinity as x goes to infinity. You have to make the jumps smaller and smaller as x goes to infinity.

    I didn't understand your argument for (d). Can you write it out with a bit more detail?

    For (a), suppose b > 0. Then for sufficiently large x, f'(x) > b/2. More precisely, there exists an x0 such that f'(x) > b/2 for all x > x0. Therefore, for all x > x0, you have f(x) > (b/2)x + f(x0). [Why?] This means that f(x) -> infinity as x -> infinity. [Why?]

    The b < 0 case is almost identical.
     
  6. Feb 21, 2012 #5
    for b)i can create a function s.t. in some interval it is a straight line ,but when x is large enougn then f(x)=1 100<x<10000 f(x)=1/2 10000<x<!00000 f(x)=1/3 100000<x<1000000000.....

    for d) we have -epsilon<f'(x)<epsilon for x >x0 , then -epsilonx-f(x0)<f(X)<epsilonx+f(x0) since then the slope of f(X) between -epsilon and epsilon.then diveided then inequality on both side by x assume x >0 then -epsilon-f(x0)/x<f(X)<epsilon+f(X0)/x then as x->infinity then f(X0)/x=0 so we get f(X)/x convergent to 0 done
     
  7. Feb 21, 2012 #6

    jbunniii

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    Yes, that works. Or, if you wanted a function that is easier to express, you could use something like f(x) = 1/[x].

    Actually it would be [itex]-\epsilon x + f(x_0) < f(x) < \epsilon x + f(x_0)[/itex] (plus, not minus, in the first expression).

    The rest looks OK.

    You should probably provide some justification for how you transformed

    [tex]-\epsilon < f'(x) < \epsilon[/tex]

    into

    [tex]-\epsilon x + f(x_0) < f(x) < \epsilon x + f(x_0)[/tex]

    i.e. Did you use the fundamental theorem of calculus? And if so, what happens if [itex]f'[/itex] is not integrable?
     
  8. Feb 21, 2012 #7


    no, but anyway thank you for helping me to solve the problem
     
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