This has it in terms of delta. How can we define delta so that 2 delta is essentially going to be epsilon? This is our chance. We will just define, we will make, and this is where we're defining delta as a function of epsilon. We're going to make 2 delta equal epsilon, or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. If you make delta equal epsilon over 2, if you make, let me switch colors just to ease the monotony, if you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2 is just going to be less than epsilon. This is the key.