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Math Help - Trigonometry and continuity

  1. #1
    Senior Member I-Think's Avatar
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    Apr 2009
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    Trigonometry and continuity

    Prove f(x)=\frac{sin x}{1+x^2} is uniformly continuous over R

    Stuck on this one

    So I'm trying to show for appropiate \delta, and |x-a|<\delta, that |f(x)-f(a)|<\epsilon

    So
    |\frac{sin x}{1+x^2}-\frac{sin a}{1+a^2}|<br />

    =|\frac{sinx-sina+a^2sinx-x^2sina}{(1+x^2)(1+a^2)}|<br />

    Known that |x-a|\geq{|sinx-sina|}

    Using triangle inequality and considering the numerator only
    |sinx-sina+a^2sinx-x^2sina|\leq{|x-a|+ |a^2sinx-x^2sina|}<br />

    And here I'm stuck.
    Hints please and thank you
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  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
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    You could instead try to show that the derivative f'(x) is bounded. If this is the case, then by the Mean Value theorem, |f(x)-f(y)|=|f'(c)||x-y|\leq M|x-y|. Choosing \delta=\varepsilon/M gives |f(x)-f(y)|\leq M|x-y|<M(\varepsilon/M)=\varepsilon, and we have uniform convergence.

    Try to show that f'(x) is bounded.
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