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Math Help - Prove continuity of a fuction

  1. #1
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    Smile Prove continuity of a fuction

    Prove that f(x)={xsin(1/x), if x≠0
    { 0 , if x=0 is continuous using

    i) language of limits
    ii) ε - δ - language

    Please help me
    Thanks in advance for any replies.
    Feyomi.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by feyomi View Post
    Prove that f(x)={xsin(1/x), if x≠0
    { 0 , if x=0 is continuous using

    i) language of limits
    ii) ε - δ - language

    Please help me
    Thanks in advance for any replies.
    Feyomi.
    clearly each piece of the function is continuous when x \ne 0. to show that f(x) is continuous then, we need only prove that \lim_{x \to 0}f(x) = f(0) = 0 (this si by the definition of continuity at a point).

    for (i) use the squeeze theorem to show this. that part shouldn't be that hard.

    for (ii) apply the \epsilon - \delta definition of the limit. do you know what it is?
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    ah yes, i understand the question much more now.
    However, is it enough the show that it is continuous at x=0?
    Would I not need to prove that it is continuous otherwise?

    I have just applied the Sandwich Rule (as I like to call it), but I do not know how to do the second part.
    I do know the definition, but I don't know how to apply it..
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    Quote Originally Posted by feyomi View Post
    ah yes, i understand the question much more now.
    However, is it enough the show that it is continuous at x=0?
    Would I not need to prove that it is continuous otherwise?

    I have just applied the Sandwich Rule (as I like to call it), but I do not know how to do the second part.
    I do know the definition, but I don't know how to apply it..
    to prove the function is continous, we need to prove it is continuous at all points in its domain. now, for x \ne 0, the function is a product of two functions that are continuous when x \ne 0 and hence the function is continuous when x \ne 0. the only problem is when x = 0, and so that's what we have to deal with.

    as for part (ii), i will give you a hint. \bigg| x \sin \frac 1x - 0 \bigg| = \bigg| x \sin \frac 1x \bigg| \le |x| for x \ne 0

    what can you do with that?
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    Quote Originally Posted by Jhevon View Post
    to prove the function is continous, we need to prove it is continuous at all points in its domain. now, for x \ne 0, the function is a product of two functions that are continuous when x \ne 0 and hence the function is continuous when x \ne 0. the only problem is when x = 0, and so that's what we have to deal with.

    as for part (ii), i will give you a hint. \bigg| x \sin \frac 1x - 0 \bigg| = \bigg| x \sin \frac 1x \bigg| \le |x| for x \ne 0

    what can you do with that?
    |xsin(1/x)| = |f(x)-0| ≤ |x| = |x-0|
    I would need to show that |x|<δ implies |f(x)|<ε ?
    I'm sorry, Sir, I'm quite clueless as to what to do
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    Quote Originally Posted by feyomi View Post
    |xsin(1/x)| = |f(x)-0| ≤ |x| = |x-0|
    I would need to show that |x|<δ implies |f(x)|<ε ?
    I'm sorry, Sir, I'm quite clueless as to what to do
    don't worry about being lost. it was difficult for all of us the first few times.

    we want to choose a \delta > 0 so that |f(x) - 0|< \epsilon (for any \epsilon > 0) whenever |x| < \delta

    so what would you choose \delta to be?
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    δ = ε/sin(1/x) ?
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    Quote Originally Posted by feyomi View Post
    δ = ε/sin(1/x) ?
    hehe, no. you can't have x in it. it must be some constant dependent on \epsilon

    look carefully at what i have so far, it's right there. i gave you what we can bound |f(x) - 0| with (and sin(1/x) doesn't come into it)
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    Yeah I just don't know..
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    Quote Originally Posted by feyomi View Post
    Yeah I just don't know..
    look at what we want.

    we want |f(x) - 0| = \bigg| x \sin \frac 1x \bigg| \le |x| < \epsilon

    and we also want |x| < \delta

    you can't see a convenient \delta to choose?
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  11. #11
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    Oh okay so we take δ=ε ?
    And then is that the end of the proof? Seeing as we've found a value of δ such that |x-0|<δ implies |f(x)-0|<ε ?
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    Quote Originally Posted by feyomi View Post
    Oh okay so we take δ=ε ?
    And then is that the end of the proof? Seeing as we've found a value of δ such that |x-0|<δ implies |f(x)-0|<ε ?
    Yes! Bingo!

    Let \epsilon > 0 be given and choose \delta = \epsilon. Then, 0 < |x - 0| = |x| < \delta implies

    \bigg| x \sin \frac 1x - 0 \bigg| = \bigg| x \sin \frac 1x \bigg| \le |x| < \epsilon

    Thus, \lim_{x \to 0} f(x) = 0 = f(0) as desired.
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    We got there in the end haha.
    Your help's been much appreciated. Thank you.
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