Formal definition of limits Part 4: using the definition | AP Calculus AB | Khan Academy
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when asking about or explaining how epsilon-delta proofs work in practice, especially how delta is chosen from epsilon.
Recommended by 1 people across 1 different months, 2014 to 2022.
What it is
Sal Khan works through a concrete epsilon-delta proof of a limit, showing how to find delta as a function of epsilon so that |x-c|<delta guarantees |f(x)-L|<epsilon. It is part 4 of a Khan Academy series on the formal definition of limits, shown as a handwritten whiteboard walkthrough.
What people said
Recommends kiratpreet dhillon · 2022-09 · Mathematics Stack Exchange · on “Epsilon delta with point discontinuity?”
The original question was posted long time ago. However, I had the same doubt. I think I understand it a little better after some research. So the Epsilon-delta definition states we can find a delta around x to satisfy that if |x-a|< delta then |f(x)-L|< any small number (Epsilon). Here f(x) doesn't necessarily need to be evaluated at "a". The definition such state that f(x) evaluated at all values at |x-a|< delta. Meaning delta is always >0 and cannot be equal to zero. here's a nice video explaining that: [this video]
Cites user170715 · 2014-05 · Mathematics Stack Exchange · on “Epsilon delta proof”
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 ( [this video] ), 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
Cites tryingtobeastoic · 2021-09 · Mathematics Stack Exchange · on “Question about the formal definition of limits”
Here [this video] Salman Khan discusses about the formal definition of limits. He tries to prove that a function $f(x)$ has a limit $L$ if $x$ approaches $c$ by expressing $\delta$ as a function of $\epsilon$ . In other words, if $\epsilon$ is given, if we can find the value of $\delta$ so that all the $x$ values that satisfy $|x-c|<\delta$ , also satisfy $|f(x)-L|<\epsilon$ , the limit exists. My questions: Is my description of the formal definition of limits correct? Can we go the other way around? I mean that if we are given $\delta$ , if we can find the value of $\epsilon$ so that all $x$ values that satisfy $|f(x)-L|<\epsilon$ , also satisfy $|x-c|<\delta$ , can we say that the limit exists?