This immediately reminded me to when I was thought the limit of 1/x as x goes to 0 to illustrate why one can't divide by 0.

but if we use L Hospital rule and differentiate the numerator and denominator then we have 1/1-0 which equals 1 so by that method the limit should exist and should be equal to one

@@user-bd1tk1yr1w Not familiar with the L Hospital rule, So I wonder how you got from x/(x-1) to 1/(1-0) or 1/1-0.

​​@@user-bd1tk1yr1w We can only apply the L'Hospital's rule if the direct substitution returns an indeterminate form, that means 0/0 or ±∞/±∞. As bprp said, this is *NOT* an undeterminate form. The limit is clearly ∞, but depending on the direction from which we approach 1, the sign of ∞ changes

Brilliant. I like this kind of videos. Subscribed🎉

Well, in both cases the function approaches 1/0. However, where x is less than 1 and greater than 0, the function is negative. Likewise, when x is greater than 1 or less than 0 (which is irrelevant to this question), the function is positive.

I did learn about that (albeit 25 years ago, oh god, and not with this guy’s notation) and my first reaction was that it doesn’t specify from which side it is in the problem.

@@JasperJanssen Why is that a problem? If the side is not specified, it's obviously the standard, and not a one-sided limit.

@@thetaomegatheta … did you watch the video? And no, there is no such thing as “the standard”.

@@JasperJanssen ' … did you watch the video?' Yes, I did. 'And no, there is no such thing as “the standard” Do you seriously not know about the non-one-sided limits?

I made the same mistake

same mistake here😢

Same and got answer one

If it's a school or university question, L'Hôpital's rule shouldn't be accepted as a valid proof to be honest

@@vintovkasnipera Why's that?

While i do agree that graphs are amazingly helpful,i believe more complex problems would be better suited for them,idk.. to me, the answer felt glaringly obvious from the start.

@@levaniandgiorgi2358 I largely agree. I looked at the statement and pretty much saw the answer immediately, but then I have the advantage over freshman students of having done math for close to 60 years. I’ve encouraged students to not shy away from sketching Bode plots or pole/zero diagrams in the EE courses I’ve taught. It is handy to look at a problem with more than one method to avoid mistakes. The backup method doesn’t need to be precise, just accurate enough to confirm your thinking is on track.

No reason to waste time graphing something like this

Graphs do what Graphs are supposed to do, give you a visual representation of the abstract equation It's helpful for people who are stronger visual learners to link the reasoning and the answer together

Graphs are very powerful. It's really hard to believe calculus was invented without using them. People make fun of 'trivial' stuff like rolle's theorem, but good luck proving them without graphs.

Thats how most people say videos are teaching more than school In reality its just bringing stuff back.

😂😂😂😂😂🤗🤗🤗

it feels like an old rusty gear inside your head has started to turn

I love the movie reference.

took multiple calc DQ and didn't really like it, but watch

studying in class 10 but still watched cuz why not 😂

Admirable curiousity. 👍

​@@jamescollier3 I never really liked math as a whole in college. Then once I graduated, I started learning it in depth, on my own . Then I started loving it. Now it has been 7 years since I graduated and I still learn it

My calc class explained SOOOO MANY of the questions I had about things that didn't make sense from earlier classes. Can recommend.

Sorry, I don't understand what positive and negative zero mean. Could you please explain?

​@@krishnannarayanan8819shorthand for "approaches zero from the positive direction" and negative directing, respectively

​@@krishnannarayanan8819 A shorthand way of saying to approach 0 from x < 0, or x > 0.

@@krishnannarayanan8819it means 0 or -0

@@krishnannarayanan8819 you could assume some number "h" which is a very small positive number. positive 0 means 0+h and negative 0 means 0-h . basically 0+ and 0- are approaching 0 from right and left sides respectively.

This channel is way out of my league 99% of the concepts he deals with... but I still come back to watch more, lol. I like how he explains things and how he writes on his whiteboard.

And that's how you get 0 points for the question on an exam. You're expected to do the calculus on a calculus exam.

@@No-cg9kj read more carefully

Approaching a limit doesn't require a Y-axis, you're needlessly complicating the concept and conditioning them with a euclidian bias in their thinking about numbers.

@No-cg9kj On any calculus exam, this question would be a multiple-choice or short answer question where you’re either right or wrong, no partial credit. Graphing is sometimes way faster than doing the math. just visualizing the graph i solved this problem in probably 2-3 seconds

@@iamcoolkinda 'On any calculus exam, this question would be a multiple-choice or short answer question where you’re either right or wrong, no partial credit' Literally none of the math exams that I took at university had multiple-choice questions. You needed to actually demonstrate your knowledge of the topic, and, in the case of specific problems like that one, you had to present solutions.

Glad to hear! Thank you!

Im grade 7(ph) and i understand calculus :)))))

DNE actually means "does not exist" lol, a limit can still exist even if theres no convergence at any particular point, e.g. diverging to positive infinity/ negative infinity

maybe its TNCAAPP: "theres no covergence at any point"

@@bartiii7617 limits diverging to positive or negative infinity also do not exist by the delta epsilon definition, though most people just go with the "you know what i mean" equals sign

​@@hyperpsych6483can't you use the epsilon-N/delta-M/N-M definitions for those limits?

@@hyperpsych6483I think it depends on the topology we’re working on, tho in common topology of ℝ we regard ±∞ as DNE, so I agree with you

You can have a limit equal to positive or negative infinity. Also, aside from Hopital you can use Taylor series.

@@lugia8888 limit equal to positive or negative infinity is typically considered DNE though, right? Because it approaches different values from left and right.

​@@wills4104 By definition, for a limit to exist in the first place, it must be a finite number. Both +∞ and -∞ aren't 'finite'. So when either the LHL or the RHL approaches either quantity, we say the limit doesn't exist.

Thank you! You too!

Well, that depends on how good your glances are, doesn’t it :)

@@literallyjustayoutubecomme1591 Agree. To beginning Calculus students, a limit often "looks DNE" as soon as they see that 0 in the denominator (even if the numerator also approaches 0).

As someone else stated, it sort of depends on how good your “glance.” Is. If your very proficient with limits and calculus, you potentially could have know just by looking at the limit what the answer would be. But a great example in this case would be x/(x-1)^2. Having a square term in the denominator actually causes the limit to approach positive infinity from both the left AND the right. Therefore the limit actually approach’s infinity and therefore does exist!

@@jackbrax7808 Depends on what you mean by "exist." According to most basic Calculus books I'm familiar with, if the limit is ∞, the limit doesn't exist-you're just being more specific about how/why it doesn't exist.

@@Steve_Stowers I just double checked my definitions and turns out your right. It doesn’t exist but both sides tend to infinity. But due to infinity not being a number, it doesn’t exist. But you can say the limit tends to infinity.

When I took a course on it, lecturer said it doesn't exist (rather than its infinity), but on the video he's calling it infinity and -infinity, and for that reason the limit doesn't exist. I think I prefer saying it doesn't exist, but saying its infinity or -infinity gives you more insight into the shape of the graph I guess.

@@colinjava8447 The limit exists if and only if the right limit equals the left limit. If left limit is different from right limit (in a given point), the limit does not exist. The limit is not "either inf or -inf", it just "does not exist".

@@janskala22 I know, that's how I knew in 2 seconds that it doesn't exist (cause left =/= right). My point was in the video he writes infinity, when like you said it just doesn't exist. I think he knows that probably but does it for convenience.

@@colinjava8447 He only writes infinity on the right limit when it holds. He writes -infinity on the left limit where it holds. He does not write any definitive answer to the whole limit until he is sure it's DNE.

@@janskala22 I know, I saw the video too.

Same

The space that is assumed in the video is the standard extension of R with two points at infinity - +∞ and -∞. Unsigned ∞ does not exist in that space. I think it's a bad decision on the author's part to not explicitly state what space we are looking for a limit in, as in other extensions of R that limit does exist.

Huh, that actually is a really nice way of thinking of it. Thanks for this!

It might help you with limits but it's not functionally the same. It's worth learning what 0+ actually means and sticking to that it will help you later with series etc

It's how I was taught at uni. Best to check with your professor if you want to be sure. Don't forget you can always (and often should) add annotations in plain language explaining what you're doing and why. Then it doesn't really matter what notation you use, as long as it's clearly defined and consistently applied.

​@@dielaughing73 our professor also uses these notations what the hell is wrong with it

This is how I was taught as well. If a teacher or other person who will check your work isn't fine with it, it's time to talk to their boss.

Just replace 1+ with 1+ε and 1- with 1-ε.

@@billmilligan7272thanks for your input Karen

'Man, my teacher has been hammering our class wuth this for months and bearly anyone understood even the priniples of "limit" and how you check for it' Let me guess, your teacher said that a limit is something that a function gets closer and closer to as its argument gets closer to some point? Yeah, I'd advice looking up an actual definition of a limit.

If you have a function 'f' which is defined on a subset 'M' of real numbers and you have some real number 'y', then the left-side limit 'lim_{x->y-} f(x)' is defined as the limit 'lim_{x->y} g(x)' where 'g' is the same as 'f' but restricted to the subset of 'M' containing only the numbers that are at most 'y'. The right-side limit is defined analogous.

So visually speaking, you cut your function at the point 'y' into a left side and a right side and handle each side on its own.

how did u know i nut behind the whiteboard? also he answered those as + and - infinity already.

@@marvinliraDEThe point is, the limit of the function defined on ℝ\{1} doesn't exist. The problem is in the question (asking for something that doesn't exist,) not in the answer, for DNE is not the limit of the function.

​@@clmasse I'm not sure there's anything wrong with asking a mathematical question for which there is no defined answer. Would you feel better if the question was "what is the limit of as x approaches , if such limit exists"? Because I'd think the qualification is implied for students beyond the most rudimentary level of maths.

exp(x) is factored out of the limit. The remaining limit is [exp(h)-1]/h which evaluates to 1.

@@Syndicalism nice! I looked at the standard way I guess you could call it for evaluating the last part, where you say that some variable, eg k = the top part, so it becomes k->0 k/ln(k+1). Bringing the k up top to the bottom w reciprocal, and then log power rule it becomes 1/ln(e) = 1. The main part that was sort of unexpected for me was the start, setting the top to a variable. How might we stumble upon this-just trying it out bc it’s limit approaches 0 as well? Also do you think that mathematicians found out the derivatives first and then tasked themselves with proving them?

There is a solution and the solution is that the limit does not exist.

​@@isaacbruner65 To say that the limit does not exist is just another way of saying that there is no solution to the problem of finding the limit. The statement that a solution does not exist is not in itself a solution. If it were, then you could say that every equation has a solution, which makes a nonsense of the concept of a solution.

​@@omp199it's different with limits. "The limit does not exist" is part of the set of possible solution with limits. Similar to NaN is a possible answer to a floating point operation even though NaN literally means Not a Number.

@@kazedcat No. It's not "different with limits". A solution is a value or set of values that satisfy a given set of conditions. If the condition is that of being the limit of an expression, then the nonexistence of a limit implies the nonexistence of a solution. As for "Nan", you are bringing programming language conventions into a discussion of mathematics. A programming language might have a function that returns NaN in certain circumstances, but that has nothing to do with mathematics.

@@omp199 Programming is mathematics. The Turing Machine is a mathematical object.

Vertical asymptote?

If we just assume the standard extension of R with two points at infinity, then yes. If we don't, there is another fairly standard space where the limit does exist - the standard extension of R with one point at infinity.

f is not odd at all.

f' (x) := lim (h->0) (f(x+h) - f(x))/h well that's how I was taught how to find the derivative, using First Principles as it were. However when they teach Calc1 now they miss out this and expect you to look the derivative up in a table, usually supplied with the exam, whats the point of even learning calc this way! 😟

1/0 is not allowed. 1+ and 1- are not numbers. 1/x has a singularity at x=0. This singularity is shifted in his example.

We don’t have an indeterminate form.

Because L’hôpital rule only works with 0/0 or infinity/infinity

Understood

Because the injury isn't bad enough to go to L'Hospital.

@@teelo12000 This isn’t ‘la Páris’! 😂😂😂😂😂😂😂😂😂😂😂😂

That would imply that it tends to positive infinity which is obviously not the case.

@@isaacbruner65 🤓

@@isaacbruner65 No, it would not. There are multiple extensions of R, one of which is assumed by BPRP in the video and has two points at infinity, and another one has only one point at infinity, which we can call 'unsigned infinity' for clarity's sake. In the case of the latter one, the person you responded to is absolutely correct, and it is a bad thing that BPRP did not explicitly bring up the matter of the space in which we are supposed to look for a limit.

@@lucaspanto9650 Your dumbass said the limit tends to infinity. I doubt you're in a position to use that emoji.

Do you want a sticker?

@@General12th lol

Thanks! Merry Christmas to you as well!

No. Try it yourself and see.

Yes. This is why the limit doesn't exist, because you get conflicting answers, depending on how you approach it. In order for the limit to exist, all possible approaches have to yield the same result. For 1/x^2 when limited to the real numbers, both possible approaches yield the same result. But, if you account for complex number approaches, you get conflicting answers, and the limit doesn't exist.

'it is ±infinite depends on which side you take' lim(x/(x-1)) as x->1 considers points in the entire neighbourhood of 1. You are thinking of one-sided limits.

Well, every neighbourhood of 1 in the real line contains elements that are less than 1 but are greater than 0. Specifically, 0.99999 is less than 1, but is greater than 0.

Imagine a number very slightly lower than 1. Such as 0.9999. That's still greater than 0.