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

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

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

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

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


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

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

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

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


    So if we let $\displaystyle \displaystyle \begin{align*} \delta = \min\left\{ 1, \frac{3\epsilon}{11} \right\} \end{align*}$ 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:
    $\displaystyle \left| {\frac{{(x - 1)(2x + 1)}}{{x + 3}}} \right| = \left| {\frac{{2x + 1}}{{x + 3}}} \right|\left| {x - 1} \right|$

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

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

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

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

    Pick $\displaystyle \delta = \min \left\{ {1,\frac{{3\varepsilon }}{5}} \right\}~.$
    Last edited by Plato; Sep 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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