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Math Help - Real Analysis Proof

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    Real Analysis Proof

    Please can someone help me, I'm hopeless with proofs and this is driving me crazy.

    If lim_{x \to a} g(x) = m with m \neq 0 then
    lim_{x \to a} \frac{1}{g(x)} = \frac{1}{m}
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    Quote Originally Posted by Nightfly View Post
    Please can someone help me, I'm hopeless with proofs and this is driving me crazy.

    If \lim_{x \to a} g(x) = m with m \neq 0 then
    \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{m}
    There are two cases here is the first. Suppose m > 0

    Since the limit exits \delta_1 > 0 such that when |x-a|< \delta_1 \implies |g(x)-m|< \frac{m}{2} \implies

    -\frac{m}{2}< g(x)-m < \frac{m}{2} \implies \frac{m}{2} < g(x)

    you can show something similar when m <0 together this implies that

    Now choose a \delta_2 such that when

    |x-a|< \delta_2 \implies |g(x)-m|< \frac{m^2 \epsilon}{2}

    let \delta =min\{\delta_1,\delta_2 \}

    if |x-a|< \delta

    |\frac{1}{g(x)}-\frac{1}{m}|=\frac{|m-g(x)|}{|m\cdot g(x)|}<\frac{\delta}{m\frac{m}{2}}< \epsilon
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