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Math Help - finding negative number N so that f(x) - L <e

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    finding negative number N so that f(x) - L <e

    A positive number e and the limit L of a function f at -infinity are given. Find a negative number N such that ABS(f(x) -L) <e if x<N

    lim
    x approaches - infinty of 1/(x+19) =0
    e= .008

    N=?
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    Re: finding negative number N so that f(x) - L <e

    Quote Originally Posted by zalar81957 View Post
    A positive number e and the limit L of a function f at -infinity are given. Find a negative number N such that ABS(f(x) -L) <e if x<N

    lim
    x approaches - infinty of 1/(x+19) =0
    e= .008

    N=?
    I can barely understand your question. Are you asking to prove \displaystyle \begin{align*} \lim_{x \to -\infty}\frac{1}{x + 19} = 0 \end{align*}? If so, you need to show that \displaystyle \begin{align*} x < N \implies \left|f(x) - L \right| < \epsilon \end{align*} for some \displaystyle \begin{align*} N < 0, \epsilon > 0 \end{align*}.

    \displaystyle \begin{align*} \left|f(x) - L \right| &< \epsilon \\ \left|\frac{1}{x + 19} - 0 \right| &< \epsilon \\ \left|\frac{1}{x + 19}\right| &< \epsilon \\ \frac{1}{|x + 19|} &< \epsilon \\ |x + 19| &> \frac{1}{\epsilon} \\ |x| + |19| \geq |x + 19| &> \frac{1}{\epsilon} \textrm{ by the Triangle Inequality} \\ |x| + 19 &> \frac{1}{\epsilon} \\ |x| &> \frac{1}{\epsilon} - 19 \\ x < -\left(\frac{1}{\epsilon} - 19\right) \textrm{ or } x &> \frac{1}{\epsilon} - 19 \\ \textrm{So we can choose to have } x &< 19 - \frac{1}{\epsilon} \end{align*}

    So we can choose \displaystyle \begin{align*} N = 19 - \frac{1}{\epsilon} \end{align*} and reversing each step will prove \displaystyle \begin{align*} x < N \implies \left|f(x) - L\right| < \epsilon \end{align*}
    Last edited by Prove It; August 1st 2012 at 08:14 AM.
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