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Math Help - Differentiability

  1. #1
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    Differentiability

    f(x) = x sin(1/x) for every x not equal to 0
    0 for x = 0

    How do i show whether this function is differentiable or not at x=0. First of all i know that if a function is differentiable at a then lim(h->0) (f(a+h)-f(a))/h has to exist or rather lim(x->a) (f(x)-f(a))/(x-a). I was pretty sure that the function was not differentiable at x=0 until i plugged it into my graphics calc to have a look at the graph, need an explanation for this please.
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    nobody wanna help me? all i know that if a function is differentiable at point a then the limit lim(h->0) (f(a+h)-f(a))/h exists but i tried doing this with some random function that is obviously not differentiable at some point and yet i managed to prove that it is differentiable which obviously means that i have done something wrong. so can somebody please give me a hand showing f(x) from the previous post (piecewise function btw) is diff or not diff at x=0. any help is appreciated.
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  3. #3
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    Quote Originally Posted by ah-bee View Post
    nobody wanna help me? all i know that if a function is differentiable at point a then the limit lim(h->0) (f(a+h)-f(a))/h exists but i tried doing this with some random function that is obviously not differentiable at some point and yet i managed to prove that it is differentiable which obviously means that i have done something wrong. so can somebody please give me a hand showing f(x) from the previous post (piecewise function btw) is diff or not diff at x=0. any help is appreciated.

    It is not differentiable at x=0.
    <br />
\lim_{h\to 0} \frac{f(h) - f(0)}{h} = \lim_{h\to 0} \frac{h\sin \frac1{h} - 0}{h} = \lim_{h\to 0} \sin \frac1{h} = \text{ Does not Exist }
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    thats what i thought, now for another 2 similar questions, can u please go through and do it to clarify my thinking process for me

    f(x) = x^2 sin(1/x) for x not equal to 0
    0 for x = 0

    g(x) = x^2 sin(1/x) for x not equal to 0
    (2/(pi)) for x = 0

    g(x) is the one that is confusing for me..
    Last edited by ah-bee; May 20th 2008 at 04:31 AM.
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    Quote Originally Posted by ah-bee View Post
    thats what i thought, now for another 2 similar questions, can u please go through and do it to clarify my thinking process for me

    f(x) = x^2 sin(1/x) for x not equal to 0
    0 for x = 0

    g(x) = x^2 sin(1/x) for x not equal to 0
    (2/(pi)) for x = 0

    g(x) is the one that is confusing for me..
    Actually f is easy, tell me the answer for f and why it is so... meanwhile I will work out g

    EDIT:Actually g is even easier, follow the same steps I did earlier and tell me where you get stuck...
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    well f is certainly differentiable because the limit is equal to 0. but the problem is, when i use the same logic with g(x) i also get a limit which is equal to 1 but i know that it is not differentiable at x=0. thats what is confusing for me.

    EDIT: i used the lim(x->a) f(x)-f(a)/x-a method for finding the limit. doubt it matters though
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  7. #7
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    btw is it right for me to be taking lim(h->0) h sin (1/h) to be zero coz the logic that im using is that -1<sin(k)<1 and h gets really really small hence it will converge to 0 right but it will be an oscillating convergence.
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  8. #8
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    Quote Originally Posted by ah-bee View Post
    btw is it right for me to be taking lim(h->0) h sin (1/h) to be zero coz the logic that im using is that -1<sin(k)<1 and h gets really really small hence it will converge to 0 right but it will be an oscillating convergence.
    Yes, and if you want a formal argument for that intuition you wrote try sandwiching h\sin \frac1{h} between h and -h. Since both h and -h goes to 0 as h goes to 0, we are through.

    Quote Originally Posted by ah-bee View Post
    well f is certainly differentiable because the limit is equal to 0. but the problem is, when i use the same logic with g(x) i also get a limit which is equal to 1 but i know that it is not differentiable at x=0. thats what is confusing for me.

    EDIT: i used the lim(x->a) f(x)-f(a)/x-a method for finding the limit. doubt it matters though
    How can you get a limit of 1? The limit does not exist actually....

    <br />
\lim_{h\to 0} \frac{f(h) - f(0)}{h} = \lim_{h\to 0} \frac{h^2\sin \frac1{h} - \frac{2}{\pi}}{h} =0 - \lim_{h\to 0}\frac{2}{h\pi}  = \text{ Does not Exist }<br />
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    so ive got

    lim(x->0) (x^2 sin(1/x) - (2/pi))/(x-(2/pi))
    simplifies to

    lim(x->0) x^2 sin(1/x)/(-2/pi) + lim(x->0) (2/pi)/(2/pi)

    which gives me 1

    edit: I am an idiot. IT WAS MY MISTAKE, sorry to waste your time lolz. i really appreciate your help, u have helped me so much. THANKS
    Last edited by ah-bee; May 20th 2008 at 05:12 AM.
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  10. #10
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    Quote Originally Posted by ah-bee View Post
    so ive got

    lim(x->0) (x^2 sin(1/x) - (2/pi))/(x-(2/pi))
    simplifies to

    lim(x->0) x^2 sin(1/x)/(-2/pi) + lim(x->0) (2/pi)/(2/pi)

    which gives me 1

    edit: sorry about the messiness, would be much more neater if i knew how to use the notation stuff that u are using. not sure if it is just an algebraic mistake or something which im pretty sure it isnt, i just did some manipulation using the limit laws.
    lim(x->0) (x^2 sin(1/x) - (2/pi))/(x-(2/pi)) is wrong.

    f(0) = 2/pi, and not f(2/pi) = 2/pi
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