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Math Help - Derivatives, differentiabilty, continuity, boundedness

  1. #1
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    Derivatives, differentiabilty, continuity, boundedness

    Let g(x) =x^asin(1/x) if x is not 0
    g(x)=0 if x=0

    Find a particular value for a such that
    a) g is differentiable on R but such that g' is unbounded on [0,1].
    b) g is differentiable on R with g' continuous but not differentiable at zero
    c) g is differentiable on R but and g' is differentiable on R, but such that g'' is not continuous at zero.


    I've only started on a) but I got a little stuck
    Well I know that a function is differentiable if g' exists for all points.
    I started by calculating derivatives:
    g'(x)=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x) if x is not zero.
    Now for x=0, we have:
    g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
    I know limxsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 provided a>2.
    So g is differentiable provided a>2.
    Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
    So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.
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  2. #2
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    Quote Originally Posted by mathematic View Post
    Let g(x) =x^asin(1/x) if x is not 0
    g(x)=0 if x=0

    Find a particular value for a such that
    a) g is differentiable on R but such that g' is unbounded on [0,1].
    b) g is differentiable on R with g' continuous but not differentiable at zero
    c) g is differentiable on R but and g' is differentiable on R, but such that g'' is not continuous at zero.


    I've only started on a) but I got a little stuck
    Well I know that a function is differentiable if g' exists for all points.
    I started by calculating derivatives:
    g'(x)=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x) if x is not zero.
    Now for x=0, we have:
    g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
    I know limxsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 provided a>2.
    So g is differentiable provided a>2.
    Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
    So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.
    Since g(x) is continuous and smooth except at x=0 it is differentiable on x \in\{ \mathbb{R}\setminus 0 \}

    Now at x=0
    \displaystyle g'(x)=\lim_{x \to 0}\frac{g(x)-g(0)}{x-0}=\lim_{x\to 0}\frac{x^a\sin\left( \frac{1}{x}\right)}{x}=\lim_{x \to 0}x^{a-1}\sin\left( \frac{1}{x}\right)

    This gives

    \displaystyle\lim_{x \to 0} -x^{a-1}\le \lim_{x \to 0}x^{a-1}\sin\left( \frac{1}{x}\right) \le \lim_{x \to 0}x^{a-1}

    This limit goes to zero if a>1. How does this help with part a)?
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  3. #3
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    Quote Originally Posted by mathematic View Post
    I know limxsin(1/x) does not exist, but...
    \displaystyle \lim_{x\to 0} x\sin(1/x)=0

    In fact even more could be said about similar limits, but I see that Mr Set has already covered that.

    CB
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    This gives

    \displaystyle\lim_{x \to 0} -x^{a-1}\le \lim_{x \to 0}x^{a-1}\sin\left( \frac{1}{x}\right) \le \lim_{x \to 0}x^{a-1}

    This limit goes to zero if a>1. How does this help with part a)?
    I don't really follow how you got that. But from that, I'm thinking a>1.
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    \displaystyle \lim_{x\to 0} x\sin(1/x)=0

    In fact even more could be said about similar limits, but I see that Mr Set has already covered that.

    CB
    Oh I see where I went wrong with that statement. I was thinking if we had f(x)=xsin(1/x) and f(0)=0, then f'(0) DNE.
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  6. #6
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    g' unbounded means we want for M>0, there exists x0 in [0,1] such that |g'(x0)|>M
    We had g'(x) = ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)
    Now we want |ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)|>M
    We know limx^(a-1)sin(1/x)=0, so we are only concerned with -x^(a-2)cos(1/x)
    So to be unbounded, we want a>2?
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