I know that the point of the proof is to show that you can get within $\epsilon$ of the limit, by giving a value that is within $\delta$ of $x$. But when solving for $\delta$ in terms of $\epsilon$ how is it possible to rewrite $|f(x) - L| < \epsilon$ in terms of $|x - c| < \delta$? I thought one concerned the $y$-axis and the other concerned the $x$-axis? For example in this Khan academy video (http://youtu.be/0sCttufU-jQ), khan changes the left side the inequality concerning $\delta$ to match the left side of the inequality concerning $\epsilon$. Then he essentially says since the left sides of the inequalities match then $\delta$ is equal to $\epsilon/2$. Again I do not understand how that's possible when both inequalities concern a different axis?
1 Answer
The main idea in an epsilon-delta proof is that the image of the delta-interval on the x-axis must be contained in (note: not necessarily equal to) the epsilon-interval on the y-axis. Moreover, the limit definition says "for every epsilon there is a delta..." -- think of it as like a two-player game, you have an epsilon-player and a delta-player. The epsilon-player goes first, giving a value for epsilon (so specifying a neighborhood around L on the y-axis). The delta-player needs to respond with a value for delta (neighborhood around c on the x-axis) so that every x in the neighborhood maps (by f) into the y-axis neighborhood the epsilon-player picked. If the epsilon-player can find a value for which the delta player has no response, then epsilon wins and the limit is not valid. If, on the other hand, the delta-player can find a formula which gives a valid response to every value the epsilon player could give, then the delta player has a strategy to keep the game going no matter what, and the limit is valid. So the delta-player's goal is to find such a formula. That's what the procedure you described does.
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$\begingroup$ Thank you for replying I spent more time with the material yesterday after getting you feedback and it was very helpful. However, I still do not feel as though I have completely grasped this concept and I was wondering if you could confirm my logic and determine if I am on the right track. {Due to length issue it will follow in separate comment} $\endgroup$ May 13, 2014 at 13:37
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$\begingroup$ I will use this as a reference problem {Limit of 2x-4 as x -> 2 is 0} 1) Fill in epsilon delta definition with info provided 2) Solve for epsilon in terms of distance from x{which is done by rewriting f(x) in terms of x, for example |f(x) - L| becomes |2x-4|(which is in terms of x)} 3) Isolate the distance from x, not the x itself(another thing that confused me in the beginning); which is then |x-2| < epsilon/2. This essentially means that the independent variable x must less than epsilon/2 of 2. In order to get an f(x) value that is less than epsilon of the limit. $\endgroup$ May 13, 2014 at 13:37
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$\begingroup$ Yes, you have the right idea here (and linear functions like 2x-4 are particularly straightforward in this regard) -- now that you have the expression in the form |x-c|< epsilon/2, the formula delta=epsilon/2 gives the delta-player the winning strategy. $\endgroup$ May 13, 2014 at 14:28