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Thread: 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 $\displaystyle f(x)=\frac{sin x}{1+x^2}$ is uniformly continuous over R

    Stuck on this one

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

    So
    $\displaystyle |\frac{sin x}{1+x^2}-\frac{sin a}{1+a^2}|
    $

    $\displaystyle =|\frac{sinx-sina+a^2sinx-x^2sina}{(1+x^2)(1+a^2)}|
    $

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

    Using triangle inequality and considering the numerator only
    $\displaystyle |sinx-sina+a^2sinx-x^2sina|\leq{|x-a|+ |a^2sinx-x^2sina|}
    $

    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
    485
    You could instead try to show that the derivative $\displaystyle f'(x)$ is bounded. If this is the case, then by the Mean Value theorem, $\displaystyle |f(x)-f(y)|=|f'(c)||x-y|\leq M|x-y|$. Choosing $\displaystyle \delta=\varepsilon/M$ gives $\displaystyle |f(x)-f(y)|\leq M|x-y|<M(\varepsilon/M)=\varepsilon$, and we have uniform convergence.

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