Talk:Calculus/Choosing delta

The $$ \delta $$ used to prove the existence of $$\lim_{x\rightarrow 2} x^2=4$$ is not consistent with the definition of limit. $$ \forall \epsilon > 0\ \exists\delta > 0 \quad \mbox{so} \quad \left | \ x - x_0 \right \vert < \delta\ \Rightarrow \left | \ f(x) - L \right \vert < \epsilon\ $$

The $$ \delta $$ used here is $$ \delta = \sqrt{4-\epsilon} - 2 $$

$$So \quad \sqrt{4-\epsilon} - 2 > 0 \quad \forall \epsilon>0 \begin{align} \sqrt{4-\epsilon} &> 2 \\ 4-\epsilon &> 4 \\ -\epsilon&>0 \\ \epsilon &< 0 \\ \epsilon &< 0 \\ &! \end{align} $$

The problem derives from the step where you have: $$ (\delta)\cdot(\delta + 4) = \epsilon $$ And skipped to: $$\delta=\frac{-4\pm\sqrt{16-4\cdot1\cdot\epsilon}}{2\cdot1}= -2 + \sqrt{4-\epsilon}$$ Asuming that c=ε.

However $$ (\delta)\cdot(\delta + 4) - \epsilon = 0 \therefore\ \delta=\frac{-4\pm\sqrt{16-4\cdot1\cdot-\epsilon}}{2\cdot1}=-2 + \sqrt{4+\epsilon}$$

A comment with regard to the derivation of the delta function in example 2: When solving for delta, the epsilon goes to the others side, so it gets a minus sign. (delta)^2 + 4(delta) - epsilon = 0 The delta function becomes -2 + sqrt (4 + epsilon).


 * You are Right why you didn't fix that in the main article!, you waste my time
 * i am going to change it to help the others, thanks for your good observation :)
 * i also changed a small error in
 * $$\begin{matrix}

\left| f(x) - L \right| & = & \left| x^2 - 4 \right| \\ \ & = & \left| x - 2 \right| \cdot \left| x + 2 \right| \\ \ & \le & (\delta) \cdot (\delta + 4) \\ \ & < & (\sqrt{4 + \epsilon} - 2) \cdot (\sqrt{4 + \epsilon} + 2) \\ \ & = & (\sqrt{4 + \epsilon})^2 - (2)^2 \\ \ & = & \epsilon \end{matrix}$$.
 * as it would be
 * $$\begin{matrix}

\left| f(x) - L \right| & = & \left| x^2 - 4 \right| \\ \ & = & \left| x - 2 \right| \cdot \left| x + 2 \right| \\ \ & < & (\delta) \cdot (\delta + 4) \\ \ & < & (\sqrt{4 + \epsilon} - 2) \cdot (\sqrt{4 + \epsilon} + 2) \\ \ & < & (\sqrt{4 + \epsilon})^2 - (2)^2 \\ \ & < & \epsilon \end{matrix}$$
 * I change the sign "=" in the last two lines with "<" because
 * $$\delta = \frac{-4 + \sqrt{16 - 4 \cdot 1 \cdot (- \epsilon) }}{2 \cdot 1} = -2 + \sqrt{4 + \epsilon}$$
 * so $$\epsilon$$ will stay with the sign "<" not "="
 * and also in third line i changed the sign from $$\le$$ to "<" because from the definition $$|x-2| < \delta$$$$ and |x^2 - 4| < \epsilon$$ not $$|x-2| \le \delta$$ and $$|x^2 - 4| \le \epsilon$$ as we want a range between 0 and $$\delta$$, 0 and $$\epsilon$$ not a specific value 3D Vector (talk) 00:47, 28 April 2008 (UTC)