0 divided by 0 (when should we say "undefined" vs "indeterminate") calculus basics

[@bprpcalculusbasics]

Infinity - infinity is also an indeterminate form and the answer is not always equal to 0 kzbin.info/www/bejne/lYummp2Fl9ilfrM

[@major__kong]

My daughter and I play this game at bed time. I say, "I love you". She says, "I love you, too". I say, "I love you three". She says, "I love you googolplex". I say, "I love you googolplex + infinity". She says, "I love you googolplex + infinity + Graham's Number". Do I tell her that's the same as infinity? Hahahaha 🙂

[@0LoneTech]

​​@@major__kong You're the one who started it - the exact same applied to your form with one extra term. Hers is just bigger (assuming they're the same infinity).

[@farahizzati5670]

hello sir can you do a video about finding absolute max and min value for fx=x-2cosx?

[@WhatAboutYou389]

∞ - ∞ is indeed an indeterminated form, in fact, if we see on what it is equal by this rule: a - b = c, c + b = a, and if we substitute the values, we get ∞ - ∞ = x, such that x + ∞ = ∞, now, first of all, ∞ is not a number, so we can't use him in basic operations, so the result x, is true only if we let ∞ be a number, witch it isn't, so it's incorrect, but, what if we want to know what value x can be so that x + ∞ = ∞? it's pretty easy, in fact, every number + ∞ = ∞, but not negative ∞, so let's re-write the expression: ∞ - ∞ = x, ∀x ∈ U \ {-∞} ⟹ x + ∞ = ∞, and not only we can use this methot, by seeing what x can be a result, we can use this methot with: (-∞) + ∞ or (-∞) - (-∞), wich is equal to x, ∀x ∈ U \ {+∞}

[@حامدعبدالفتاح-ت1ض]

Please, I want you to create a formula to solve the equation ax^3+bx^2+cx+b=0 And this one too ax^3+bx=n

[@zachansen8293]

Thanks for making this followup. There sure were a lot of people in the other video who were dead sure you were just wrong.

[@Pure_Calculus_MATHS]

Hi, I'm from France and I'm super interested in mathematics! I have several questions, could you solve the following integrals: the integral of |((sin(x))/(x))| and the integral of 1/(1+x^(5)). I know these are big integrals but it would be nice to do them! PS: can you make a complete guide for the 3rd year of calculus with your McDonald's employee outfit? I don't have more to say! Bye! :D

[@joe_z]

I first heard the lone word "indeterminate" being used for indeterminate forms during middle school in our introductory algebra class. It didn't make sense to me back then either because I knew that it was still a type of "undefined". I'd wager that the vast majority of people who insist that 0/0 is _not_ "undefined" have not really gone into the kind of college-level math where indeterminate forms are actually used yet.

[@teelo12000]

So which one is unstoppable force striking immovable object?

[@doctorwhouse3881]

There's systems where 0/0 can be defined as the entire set, but in most day to day systems you want operations to result in single numbers.

[@lily_littleangel]

What about the limit of x -> 0 of (x / 0); is that indeterminate or undefined?

[@kazedcat]

Undefined. Division by zero is not allowed even in Calculus. Your limit has exactly zero in the dinaminator so it is undefined.

[@حامدعبدالفتاح-ت1ض]

Please, I want you to create a formula to solve the equation ax^3+bx^2+cx+b=0 And this one too ax^3+bx=n

[@teelo12000]

Proof that 0/0=0: Start with 0*0=0, now divide by zero on both sides: 0*0/0 = 0/0 -> 0 = 0/0 Proof Complete.

[@packered]

Same Logic: 5*0=0, divide both sides by zero, 5*0/0=0/0 -> 5 = 0/0. Since this process works for all numbers, 0/0 is undefined.

[@Jee.Aspirant.]

Proof that 0/0 is 1 let 0 = x we know that x/x = 1 0/0 = 1 👽👽👽

[@blackdecker7327]

Why couldn't you do the trick even for the first problem? I mean the algebraic work.

[@Ninja20704]

We always have to go off the original definition of the function. So even though we can cancel out the x-3, we are not allowed to plug in x=3 based on the original definition of the function. Graphically, you get a smooth curve but with a hole/removable discontinuity at x=3 instead of a vertical asymptote.

[@ZipplyZane]

Because 3-3= 0, and you can't divide by 0. When doing the limit, x is not exactly 3, so x-3 is not exactly 0. So you can still divide by it.

[@Nunnayadambiness]

The answer is just simply "yes".