limits with epsilon-delta definition! (linear, square root, and quadratic examples)

Here's the easy way to write epsilon-delta proofs for limits. Usually, the epsilon-delta definition is taught in college calculus 1 (or real analysis in upper-division pure math but not in AP calculus). I think it is the hardest topic in Calculus 1. Let me know what you think!

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0:00 proving limit of 1/2x+3 as x goes to 4
3:35 proving limit of sqrt(10-3x) as x goes to -2
8:16 proving limit of 3x^2+1 as x goes to 2

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😊 Thanks to all calculus supporters 😊
Sandglass Dªrksun Seth Morris

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[nyinakuemmanuel]

I've been watching you videos since high school. I'm a Ghanaian offering BSc Statistics at University of Cape Coast.
Am in my three-year. Thanks very much red pen black pen. This semester I'll do Advanced Calculus.
So I'll watching every single of your videos on epsilon _delta proofs🙏🙏🙏👌💗

[zyriv]

As a freshman in college, when I first started studying real analysis, I got confused by how to choose correctly ε in relation with δ. But this definitely helped me, thanks !

[bautistaromaniuk]

Thank you for the vids! They are super helpful. However, I still do not understand why we can simply ignore the √10-3x +4

[jujoropo]

Wooow thank you very much! You help so much! Love your videos

[kingbeauregard]

The part about arbitrarily choosing the delta has always been the most confusing part for me, because it seems like a nonsensical step. Here's my concept of what it's about. When you've got that expression on the left and epsilon on the right, what you're trying to do is come up with increasingly simpler expressions that are always bigger* than your original function; if you can finally get to a bigger* simpler expression that you know converges to your limit, then so must your original function. The problem is, it's pretty rare that you can come up with a simpler expression that is bigger* for all values of x. So that's when you cheat: you say "well I know that suchandsuch expression will be bigger* if I keep x within a certain narrow region".

Like, think of y = e^x, and the limit at (0, 1). It would be nice to come up with some straight line that is always bigger* than e^x; then you could just say that, since the line converges to (0, 1), so must e^x. But there is no straight line that will do that for every value of x. So instead you cheat: you can say that, if we're limiting our x values to the region from -1 to 1, then for that region, a line with a slope of (e-1) that goes through (0, 1) will always be bigger* than the function. Once we make that tradeoff, we can proceed.

*: When I say "bigger", I mean a function that is further away from the limit than our original function for every value of x in question. It's a squishy word but it makes instinctive sense I think.

[CHEESYhairyGASH]

At 7:22, why do you ignore the integer in the denominator? In a previous video titled "how to easily write the epsilon-delta proofs for limits" you only ignored the square root part and kept the integer in the denominator (this happened at 6:33 in the other video).

Why do you use different methods in these two very similar cases???

[dominicgrew]

Thank you very much....great video as always

[kushaldey]

I need a little help in the last example, when we need to choose δ=min(1, ε/15). We could have said δ=min(1.5, __).
Then, |x-2|<1.5 => 0.5<|x+2|<3.5,
Then, 3|x²-4| = 3δ×3.5=10.5δ.
Thus we could have chosen δ=min(1.5, ε/10.5). Are we just choosing 1 instead of 1.5 because it's generally easy?

[yoyoezzijr]

For the second one, wouldnt delta = epsilon also work? because the denominator √(10-3x) + 4 is always greater than 3, so the expression is always less than |x+2| which is less than delta which is equal to epsilon

[jafecc]

What if we have two square roots adding together or subtracting one from another in the limits? Is it harder to prove using epsilon-delta definition? Please let us know.

[marylamb]

One reason why you want to "tie" the value of delta with the value of epsilon is that you want to prove the limit of the function for ALL values of x. There will always be only one value of delta that will be true for only one value of epsilon.

[onenutwonder]

I have no idea how or why the epsilon delta proofs work to show the limit is true but maybe that's why i find them to be so damn cool.

[dennism]

Can you do x^4/sin(x^2) as x tends to zero with epsilon delta?

[TheBluNitro]

why did -2 become -4 when you isolate 1/2 from the absolute value?

[judedavis]

For the delta inequalities, you don't actually need the 0 < part because we're dealing with absolutes?

[MilFiadis]

Ο παχατούρ μου είπε να σε δω. Καλά τα λες ρε μπαγάσα, μπράβο

[grupocelebremos]

but i didn't understood why you just can think that the min value of delta is 1 i mean, I know that is easy but why 1 is never the max value

[matteocilla]

i see this in my mathematical analysis course and i don’t understand anything….

[Stephen_]

Just be careful if you do not have given epsilon greater than zero