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Math Help - Please help me with this prove question

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    Please help me with this prove question

    Prove that if there is a number B such that |f(x)| <= B for all x not = 0, then lim x approaches 0 xf(x) = 0. Note, exercise 43-46 are special cases of this gerenal result.


    Special cases are:
    43. lim x approaches 0 sin (1/x) = 0;
    44. lim x approaches pi (x - pi) cos^2 [1/(x-pi)] = 0;
    45. lim x approaches 1 |x -1| sinx = 0;
    46.
    f(x) = {1 if x rational
    f(x) = {0, if x irrational then lim x approaches 0 xf(x) = 0.


    This is the hardest question in my assgnment, no one in my study group is able to do it, I really need anyone who can give me some hints on this, thanks!
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    Quote Originally Posted by 450081592 View Post
    Prove that if there is a number B such that |f(x)| <= B for all x not = 0, then lim x approaches 0 xf(x) = 0. Note, exercise 43-46 are special cases of this gerenal result.


    Special cases are:
    43. lim x approaches 0 sin (1/x) = 0;
    44. lim x approaches pi (x - pi) cos^2 [1/(x-pi)] = 0;
    45. lim x approaches 1 |x -1| sinx = 0;
    46.
    f(x) = {1 if x rational
    f(x) = {0, if x irrational then lim x approaches 0 xf(x) = 0.


    This is the hardest question in my assgnment, no one in my study group is able to do it, I really need anyone who can give me some hints on this, thanks!

    0 <= |xf(x)| = |x||f(x)| <= |x|B --> 0 when x --> 0, by arithmetic of limits so the squeezing (or the sandwich) lemma gives you the result.

    Tonio
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  3. #3
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    Quote Originally Posted by 450081592 View Post
    Prove that if there is a number B such that |f(x)| <= B for all x not = 0, then lim x approaches 0 xf(x) = 0. Note, exercise 43-46 are special cases of this gerenal result.


    Special cases are:
    43. lim x approaches 0 sin (1/x) = 0;
    44. lim x approaches pi (x - pi) cos^2 [1/(x-pi)] = 0;
    45. lim x approaches 1 |x -1| sinx = 0;
    46.
    f(x) = {1 if x rational
    f(x) = {0, if x irrational then lim x approaches 0 xf(x) = 0.


    This is the hardest question in my assgnment, no one in my study group is able to do it, I really need anyone who can give me some hints on this, thanks!
    45.
    0 \leq \sin{x} \leq 1

    0|x - 1| \leq |x - 1|\sin{x} \leq 1|x - 1|

    0 \leq |x - 1|\sin{x} \leq |x - 1|

    Now take the limit as x \to 1.
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    Quote Originally Posted by tonio View Post
    0 <= |xf(x)| = |x||f(x)| <= |x|B --> 0 when x --> 0, by arithmetic of limits so the squeezing (or the sandwich) lemma gives you the result.

    Tonio

    could you tell me more about it, cause we didn;t learn the lemma therom yet, how do I apply it to this proof?
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    Quote Originally Posted by Prove It View Post
    45.
    0 \leq \sin{x} \leq 1

    0|x - 1| \leq |x - 1|\sin{x} \leq 1|x - 1|

    0 \leq |x - 1|\sin{x} \leq |x - 1|

    Now take the limit as x \to 1.

    43 to 46 is not the questions, they are the result that are already proven, but thanks though
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    Quote Originally Posted by 450081592 View Post
    could you tell me more about it, cause we didn;t learn the lemma therom yet, how do I apply it to this proof?

    The sandwich lemma: let f,g,h be three functions defined in some neighborhood of a point w (excetp, perhaps, in w itself) and s.t. g(x) <= f(x) <= h(x) for all x in this neighborhood. If g(x) --> L and h(x) --> L as x --> w, then also f(x) --> L as x --> w.

    There's a corresponding very wll known lemma in limits of sequences, too.

    Tonio
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