Epsilon delta proof

Question

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?

Answer 1

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.