Where were you 20 years ago when I was lost learning this for the first time

Just wanted to quickly say thanks for this! I just had my Calc & Analysis exam today, and if I didn't watch this video last night, I wouldn't have realized that I had to use the delta = min{1,n*eplison} thing! You saved me from losing marks! Cheers from Canada!🇨🇦

Can you please do a third part and prove limits of trig functions or in general, anything other than polynomials? Great video btw, extremely helpful for my test.

I'm 38 years old and, on a whim, I decided to self teach myself all the math high school failed to teach me as a kid. When I finally got to epsilon-delta proofs for limits... well... I just gave up, called limits done, then moved on to derivatives with that huge gap in my knowledge. How they were explained in the textbooks I managed to dig up were so dense and obtuse, couldn't make heads or tails of it at all. It turns out I just needed a good teacher to run me through several examples until I managed to get it to click. Thank you!

My calc professor loved the triangle inequality, so much that it kinda spread to me

very talented took me many hours and 3 watching but starting to get it

Here's how my brain processes it. If you're trying to prove the limit of a function exists, it's hard to say anything definitive about most functions on their own; your best plan is to compare them to functions we already know very well. And we do know how limits work on a straight line, so if we can argue that we've got a straight line that goes through a given point (a, L), AND the straight line is also uniformly "bigger" than the function we're contemplating, then that function must converge to (a, L). By "bigger" I mean: at x-values where the value of the function is greater than L, then the y-value of the line is even greater than that. And likewise, at x-values where the value of the function is less than L, the y-value of the line is even less than that. So the manipulations we're doing with the inequality are intended to help us find a line that we know must always be "bigger" than the original function. There are two tricks we can pull to do this: 1) Replace any part of the inequality with a term that we know is always going to make the whole expression bigger, or not any smaller anyway. Like if you have |sin(x)| as a multiplier in the inequality, you can replace it with "1". Or, if you have a denominator of (x^2 + 3), just replace that denominator with "3". 2) Restrict yourself to a narrow range of x-values around the limit you're trying to prove. That way you can start saying things like, "well if I stay between a-1 and a+1, I know that suchandsuch term will never be more than 12, so let's replace that term with 12". So when all is said and done, if all goes well, you will be able to derive a relation between delta and epsilon that essentially describes a line that is "bigger" than the original function, for whatever narrow range of deltas we decided on in trick #2. Since you were able to derive such a line, with appropriate mathematical rigor, you can confidently say that the function must have a limit at (a, L).

Hey bprp I think it would be good if you did an undefined limit where there isn’t a solution like the stepwise function. It would be cool to see how it doesn’t work with epsilon delta.

Could you make videos on the 1/(x-1) case? i tried it and couldnt use the techinique you use with 1/x cause that would need calculating 1/0 which is impossible or tends to infinity (not useful)

can you do, lim h -> 0 of newton's difference quotient for f(x)=x^3? so it evaluates to lim h -> 0 (3x^2+3xh + h^2) = 3x^2 how to prove with delta-epsilon?

Now I have a better understanding of epsilon delta definition but I still have a doubt. In normal limit calculations we evaluate the limit. But in epsilon delta definition we need L as a requirement. We are not calculating the limit, we are just proving the limit exists. Then how is epsilon delta definition better than ordinary limit calculation?

Why we have to choose delta as min{something, epsilon}?

on my way to use 2004 instead of 1 in "min" because that's my birthyear

but x is not defined on which intervall

Can you please solve the limit of (1+3/x)^(x+2) as x->infinity?

1:46 to 5:00, I don't understand that part

Z²

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For the second half of the video --- where f(x) = 1/x --- what you explained works great for proving the lim f(x) any x such that abs(x) > Some_Positive_Real_Value where Some_Positive_Real_Value is the first part of the choose delta definition delta = min{Some_Positive_Real_Value, epsilon-something-or-other}. In your proof you had Some_Positive_Real_Value = 1. But what if the question had asked to prove the limit as x --> 1/2 ? Then this methodology would break down. I dislike the encouragement of just setting Some_Positive_Real_Value = 1 because you can easily get into trouble with abs(x) having a derived range that includes zero, which then blows up to infinity for 1/abs(x). Better to encourage some critical thinking and ask why we pick a value there of 1. Instead, for limit 1/x as x approaches c where c is in the domain of f: Domain maps to Reals, let's choose delta = min{ abs(c)/2 , epsilon-something-or-other}. By specifically bounding ourselves with a delta that is small enough to never include x=0 --- a known discontinuity for f(x) = 1/x --- we avoid some serious issues later on. This also hints at the deeper underlying reasoning behind the epsilon-delta definitions.

We are done.

Oh no, 3^2 +2*3 + 4 = 20