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Math Help - Proving the Continuity of a Function

  1. #1
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    Proving the Continuity of a Function

    Hi

    I am trying to proove that (2x^2 + 2) / (x +3) is continous at lim x -> 1 => f(x) = 1

    This is what I've thought so far

    |f(x) - 1 |< Epsilon
    |(2x^2 -x -1) / (x +3)| < Epsilon
    |(x - 1)(x + 2) / (x + 3)| < Epsilon

    Now I fail to prove that for some K Epsilon = Delta => |x -1| < Delta, thus proves Continouty.
    Please help me, I been trying for several hours... Going to try another problem to cool down.

    Thanks!
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  2. #2
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    Re: Proving the Continuity of a Function

    Quote Originally Posted by Requnix View Post
    Hi

    I am trying to proove that (2x^2 + 2) / (x +3) is continous at lim x -> 1 => f(x) = 1

    This is what I've thought so far

    |f(x) - 1 |< Epsilon
    |(2x^2 -x -1) / (x +3)| < Epsilon
    |(x - 1)(x + 2) / (x + 3)| < Epsilon

    Now I fail to prove that for some K Epsilon = Delta => |x -1| < Delta, thus proves Continouty.
    Please help me, I been trying for several hours... Going to try another problem to cool down.

    Thanks!
    For a function to be continuous at a point, it has to be defined at that point, and the limit of the function approaching that point is equal to the value of the function at that point.

    So to show \displaystyle \begin{align*} \frac{2x^2 + 2}{x + 3} \end{align*} is continuous at \displaystyle \begin{align*} x = 1 \end{align*}, you first need to show that the function is defined at \displaystyle \begin{align*} x = 1 \end{align*} (easy, \displaystyle \begin{align*} f(1) = \frac{2(1)^2 + 2}{1 + 3} = 1 \end{align*}), and then you need to show \displaystyle \begin{align*} \lim_{x \to 1}\frac{2x^2 + 2}{x + 3} = 1 \end{align*}.

    To do this, we need to prove \displaystyle \begin{align*} 0 < |x - 1| < \delta \implies \left|\frac{2x^2 + 2}{x + 3} - 1 \right| < \epsilon \end{align*}. Working on the second inequality we have

    \displaystyle \begin{align*} \left|\frac{2x^2 + 2}{x + 3} - 1 \right| &< \epsilon \\ \left|\frac{2x^2 + 2}{x + 3} - \frac{x + 3}{x + 3} \right| &< \epsilon \\ \left|\frac{2x^2 - x - 1}{x + 3}  \right| &< \epsilon \\ \left| \frac{2x^2 - 2x + x - 1}{x + 3} \right| &< \epsilon \\ \left| \frac{2x(x - 1) + 1(x - 1)}{x + 3} \right| &< \epsilon \\ \left| \frac{(x - 1)(2x + 1)}{x + 3} \right| &< \epsilon \\ |x - 1| \left| \frac{2x + 1}{x + 3} \right| &< \epsilon \\ |x - 1| \left| \frac{2x + 6 - 5}{x + 3} \right| &< \epsilon \\ |x - 1| \left| 2 - \frac{5}{x + 3} \right| &< \epsilon  \end{align*}

    Now if we restrict \displaystyle \begin{align*} |x - 1| < 1 \end{align*} (say), since we have to make the distance from x small anyway, this gives

    \displaystyle \begin{align*} |x - 1| &< 1 \\ -1 < x - 1 &< 1 \\ 0 < x &< 2 \\ 3 < x + 3 &< 5 \end{align*}


    We need an upper bound for \displaystyle \begin{align*} \left| 2 - \frac{5}{x + 3} \right| \end{align*}. Note that

    \displaystyle \begin{align*} \left|2 - \frac{5}{x + 3} \right| &\leq |2| + \left|- \frac{5}{x + 3} \right| \textrm{ by the Triangle Inequality} \\ &= 2 + \frac{5}{|x + 3|} \\ &< 2 + \frac{5}{3} \textrm{ since we restricted } |x - 1| < 1 \implies x + 3 > 3 \\ &= \frac{11}{3} \end{align*}

    Therefore, as long as we have \displaystyle \begin{align*} |x - 1| < 1 \end{align*}, we have

    \displaystyle \begin{align*} |x - 1|\left| 2 - \frac{5}{x + 3} \right| &< \epsilon \\ \frac{11}{3} |x - 1| &< \epsilon \\ |x - 1| &< \frac{3\epsilon}{11} \end{align*}


    So if we let \displaystyle \begin{align*} \delta = \min\left\{ 1, \frac{3\epsilon}{11} \right\} \end{align*} and reverse the process, you will have your proof
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    Re: Proving the Continuity of a Function

    Quote Originally Posted by Requnix View Post
    I am trying to proove that (2x^2 + 2) / (x +3) is continous at lim x -> 1 => f(x) = 1
    This is what I've thought so far
    |f(x) - 1 |< Epsilon
    |(2x^2 -x -1) / (x +3)| < Epsilon
    |(x - 1)(2x + 2) / (x + 3)| < Epsilon
    Once we get the that final line notice that:
    \left| {\frac{{(x - 1)(2x + 1)}}{{x + 3}}} \right| = \left| {\frac{{2x + 1}}{{x + 3}}} \right|\left| {x - 1} \right|

    That is important because we control the value |x-1|.

    Therefore we need to bound \left| {\frac{{2x + 1}}{{x + 3}}} \right|.

    So start with |x-1|<1 from which it follows that
    1<2x+1<5~\&~3<x+3<5.

    Now we know that \left| {\frac{{x + 2}}{{x + 3}}} \right|<\frac{5}{3}~.

    Pick \delta  = \min \left\{ {1,\frac{{3\varepsilon }}{5}} \right\}~.
    Last edited by Plato; September 30th 2012 at 03:09 PM.
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    Re: Proving the Continuity of a Function

    FYI, there's an excellent paper epsilondelta.pdf in the thread "For those that never learnt well the epsilon-delta proofs" which is one of the four "sticky" threads on this forum. Example 3 is almost identical to this one.

    - Hollywood
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  5. #5
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    Re: Proving the Continuity of a Function

    I've made an error, however your way of dealing with the |(x - 1) (x + 2) / (x + 3)| < Epsilon, has been most helpfull and enlightening. Thanks again.
    2x^2 - x -1 => (1 (+-) Sqrt(1 + 8)) / 4 Giving x = 1 V -(1/2). But whats most important is that (x -1) is still there
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