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Math Help - Continuity of quotient

  1. #1
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    Continuity of quotient

    I'm looking for some help with proving that if f and g are continuous then so is f/g. I need to use the ε and Δ definition of continuity
    ( │ x - a│< Δ implies │f(x) f(a)│< ε) to prove this.
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  2. #2
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    Quote Originally Posted by CoraGB View Post
    I'm looking for some help with proving that if f and g are continuous then so is f/g. I need to use the ε and Δ definition of continuity
    ( │ x - a│< Δ implies │f(x) f(a)│< ε) to prove this.

    Of course, we must assume g(a)\neq 0, and then :

    \forall,\epsilon>0\,\,\exists\,\delta_1>0\,\,s.t.\  ,\,|x-a|<\delta_1\,\Longrightarrow\,|f(x)-f(a)|<\frac{e}{2g(a)}\,\,\,and\, \exists\,\,\delta_2>0\,\,\,s.t.\,\,|x-a|<\delta_2\,\Longrightarrow\,|g(x)-g(a)|<\frac{2}{2f(a)}

    We also have that |g(a)-g(x)|<\frac{\epsilon}{2f(a)}\Longrightarrow\,|g(x)  |>|g(a)|-\frac{\epsilon}{2f(a)} , thus for |x-a|<min(\delta_1,\delta_2):

    \left|\frac{f(x)}{g(x)}-\frac{f(a)}{g(a)}\right|=\left|\frac{f(x)g(a)-f(a)g(a)+f(a)g(a)-f(a)g(x)}{g(x)g(a)}\right| \leq\frac{\left|f(x)-f(a)\right||g(a)|+\left|g(a)-g(x)\right||f(a)|}{|g(x)g(a)|} <\frac{\frac{\epsilon}{2g(a)}|g(a)|+\frac{\epsilon  }{2f(a)}|f(a)|}{|g(x)g(a)|}=\frac{\epsilon}{K} , where K is a

    constant (as g(x) is as close to g(a) as we want if x close enough to a).

    Tonio
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