proving the limit of a product is the product of the limits, epsilon-delta definition

We will prove the limit of a product is the product of the limits (assuming the limits exist) by using the epsilon-delta definition of a limit. This is a classic proof that you will see in your advanced calculus class or real analysis class. This is hard proof especially when you first see it. Be sure you put in the effort and study it well. #calculus #realanalysis #blackpenredpen

0:00 Review some limits and their epsilon-delta definition
2:48 The proof!

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Thank you all!

[what_a_lame_tag_system]

You know something is wrong when bprp doesn't use whiteboard for proof

[_ym_chn_]

27:05 to avoid the 0/0 case, we can simply have epsilon*|L|/[2(|L| + 1)] < epsilon*(|L| + 1)/[2(|L| + 1)] = epsilon/2, and the solution is unaffected in any way ^^

[CallMeIshmael]

This is one of the more complicated proofs of basic analysis facts and I've never really had a good intuition for the quantities that come up in the proof, so thank you for doing a good job of talking through it.

[somerapdude]

Not to kiss your behind or whatever, but this is the most helpful channel I’ve ever found when it comes to helping me with my EXTREMELY hard math class

[epsilia]

26:07 I saw some people trying to explain how we don't have to bother about the |L|=0 case in a certain way, but in the end :
We know that |L| < [L|+1 for any real number L. So |L|/(|L|+1) < 1 as a consequence. Multiplying each sides by eps/2 (> 0) guarantees us the result !
Edit : Nevermind it has been said by multiple people already, my bad

[tzovgo]

26:27 You don't have to cancel two zeros if you add 1 to |L|, proving that step rigorously!

|L| * (epsilon)/(|L|+1) < (|L| + 1) * (epsilon)/(|L|+1) = (epsilon)/2

[thatonemailbox]

1:28 This is the best explanation of the epsilon-delta definition of limits I have ever seen. Everything for me just clicked once I saw this. You're an awesome teacher!

[cgandcats]

Thank you very much!!! I Found this proof in Spivak's book but I was not getting how it was done and after watching your video now is all clear to me. Love you!

[Maths_.]

This is blackpenredpenbluepengreenpen and yellow highlighter

[mariannelee]

I'm graduating soon in Mechanical Engineering and I just have to say your channel is the best! You've helped me so much!

[jvthunder]

I remember getting this question in my university practice and getting the proof myself is so satisfying!

[martinepstein]

I prefer to use epsilon-delta arguments only in the simplest cases. For example,

- A sum of two functions approaching 0 approaches 0

- A function approaching 0 times a bounded function approaches 0

Now f can be expressed as L plus a function approaching 0, and g can be expressed as M plus a function approaching 0. You can just multiply it out and see that the product is LM plus a function approaching 0 i.e. the limit of the product is LM.

[titan]

Bprb, WHAT AN AWESOME PROOF! Even though it turned out to be a very difficult one, I understood every step of it, and that's because you are en EXCELLENT TEACHER. Thank you! 🙏🏻

[yohangross]

extremely good job and explanations. As a new online teacher, I find your content really inspiring. I now recognize that I mimic a lot your teaching style. Keep up the good job !

[thexoxob]

The thing is you see, for the initial limit isn't using the product rule for limits, it's just substituting that c value for every instance of x

[darkdelphin]

You are looking CLEAN on the thumbnail man. Nice

[scottleung]

Wow - this takes me back to my modern analysis course in college. I'll never forget how brutal that was. But good job with the proof!

[marble]

Seeing math makes me remembers when i find the answer
Like "I have 9 numbers, the first 5 numbers has a mean of 12, while the last 4 numbers has a mean of 3, what is the mean of all 9 numbers"
And thats when i realise i can just do (5 × 12 + 4 × 3) ÷ 9 and got 8

[yanceyward]

I have always loved working out these fundamental proofs. I actually did this particular one out just a few months ago working through my Schaum's Calculus workbook. I didn't think it was particularly difficult- the really key insight is that addition/subtraction trick that allows the breaking apart of the functions, but it is a common technique in algebra, so came to me almost instantly.

[srikanthtupurani]

Main trick is writing g(x) as
g(x) -M+M after this there is no need to worry about the term f(x) g(x). It is all about being careful. We have to adjust things.