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    I have shown that for x> 0, -x \leq sin(x) \leq x.

    Now I am trying to show that \int^1_0 nx sin(\frac{1}{nx}) converges to 1 as n tends to infinity, using the fact in the first line.

    From the first line, I get that -1 \leq nxsin(\frac{1}{nx}) \leq 1 but I'm not sure how to finish off to show the convergence?
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    If you know the dominated convergence theorem you can use the fact that  \lim_{x \to 0} \frac{\sin{x}}{x}=1 to get the result (you use the line above to show that the conditions of the DCT apply)

    I'm not convinced on it's own that the line above is sufficient to prove anything without invoking some heavy machinery. All you have is a trivial lower bound (we could also have used sin(x)>0 within that interval!). The upper bound is useful, but at some point you need to say more about the function sine than just than inequality (the function f(x)=0 also satisfies that inequality but would not satisfy the conclusion you are trying to achieve) I guess you could use fatou's lemma, but you still need some sort of nice theorem.
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    I think you can do it without *heavy* machinery. For x <pi/2 we have 0 <= sin(x) <= x. Integrate twice to get first cos(x) >= 1-x^2 and then sin(x) >= x-x^3/6. So x - x^3/6 <= sin(x) <= x. From there, take eps > 0 and show we can find N s.t. |nx sin(1/nx) - 1| < eps for n > N.
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    (Original post by IrrationalNumber)
    If you know the dominated convergence theorem you can use the fact that  \lim_{x \to 0} \frac{\sin{x}}{x}=1 to get the result (you use the line above to show that the conditions of the DCT apply)

    I'm not convinced on it's own that the line above is sufficient to prove anything without invoking some heavy machinery. All you have is a trivial lower bound (we could also have used sin(x)>0 within that interval!). The upper bound is useful, but at some point you need to say more about the function sine than just than inequality (the function f(x)=0 also satisfies that inequality but would not satisfy the conclusion you are trying to achieve) I guess you could use fatou's lemma, but you still need some sort of nice theorem.
    I think you are right, I think it's that I need to use the inequality in the first line of OP, and a convergence theorem to show it.

    If I used the dominated convergence theorem, could I use the inequality to show that the function fn, is always <=1 and so the dominating function could be e^x?

    Then let f(x)=1
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    (Original post by sonic7899)
    I think you are right, I think it's that I need to use the inequality in the first line of OP, and a convergence theorem to show it.

    If I used the dominated convergence theorem, could I use the inequality to show that the function fn, is always <=1 and so the dominating function could be e^x?

    Then let f(x)=1
    Bit of a weird choice of dominating function, it does work, but a more natural choice is surely the function 1.

    If you've seen it the boundedness convergence theorem also works (I only saw this because it came up a while ago on TSR- it's essentially a very simple corollary of the DCT) . DFranklin's (pretty nice) solution works but you should write somewhere (if your course is something in measure theory) that almost sure convergence implies L^1 convergence for finite measure spaces (you use this fact!)
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    (Original post by IrrationalNumber)
    Bit of a weird choice of dominating function, it does work, but a more natural choice is surely the function 1.

    If you've seen it the boundedness convergence theorem also works (I only saw this because it came up a while ago on TSR- it's essentially a very simple corollary of the DCT) . DFranklin's (pretty nice) solution works but you should write somewhere (if your course is something in measure theory) that almost sure convergence implies L^1 convergence for finite measure spaces (you use this fact!)
    Ah yep I think the bounded convergence theorem is easier to use. That way I can say that the bound is 1 and the fn(x) tend to 1 as n tends to infinity and so the integral of fn tends to the integral of 1 between 0 and 1, which is 1
 
 
 
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