Why doesn’t this limit represent the definition of derivative? 👉kzread.info/dash/bejne/dWhokteQg92sd8o.htmlsi=Zx-wuAEkbfV5_m8a

@BlacksmithTWD

5 ай бұрын

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.

@user-bd1tk1yr1w

5 ай бұрын

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

@BlacksmithTWD

5 ай бұрын

@@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.

@mondherbouazizi4433

5 ай бұрын

​​@@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

@om-qz7kp

5 ай бұрын

Brilliant. I like this kind of videos. Subscribed🎉

Honestly I think the class didn’t learn limit from right and left or they just forget about it.

@Lordmewtwo151

5 ай бұрын

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.

@JasperJanssen

5 ай бұрын

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.

@thetaomegatheta

5 ай бұрын

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

@JasperJanssen

5 ай бұрын

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

@thetaomegatheta

5 ай бұрын

@@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?

Note to self: "L'Hôpital's rule can only be applied in the case where direct substitution yields an indeterminate form, meaning 0/0 or ±∞/±∞. So if f and g are defined, L'Hôpital would be applicable only if the value of both f and g is 0. "

@homebird4765

5 ай бұрын

I made the same mistake

@vedantlearns7516

5 ай бұрын

same mistake here😢

@Sanji-ip1vd

5 ай бұрын

Same and got answer one

@vintovkasnipera

5 ай бұрын

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

@homebird4765

5 ай бұрын

@@vintovkasnipera Why's that?

This limit problem is a good illustration of why making even a rough sketch graph of the function in question can shed a lot of light. Using a graphing as a qualitative analytical tool is too often overlooked.

@levaniandgiorgi2358

5 ай бұрын

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.

@wtmayhew

5 ай бұрын

@@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.

@ivanzonic

5 ай бұрын

No reason to waste time graphing something like this

@heylolp9

5 ай бұрын

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

@sankang9425

5 ай бұрын

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.

I just woke up and this video was suggested. I haven't touched in limits for almost a decade so my thought was "Ive forgetten all of this". I've opened the video,watched for 3 minutes and I could feel the knowledge coming back 😂 so weird

@dscarmo

2 ай бұрын

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

@Maximus.Decimus

Ай бұрын

😂😂😂😂😂🤗🤗🤗

Cady Heron would have figured this out… it’s how she won the athlete competition in “Mean Girls.”

@StaticBlaster

5 ай бұрын

I love the movie reference.

I used to love asking my students on these if their denominator was positive zero or negative zero. The transition from initial confusion to a-ha was one of my favorite gems from teaching.

@krishnannarayanan8819

5 ай бұрын

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

@BlueGamingRage

5 ай бұрын

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

@l.w.paradis2108

5 ай бұрын

​@@krishnannarayanan8819 A shorthand way of saying to approach 0 from x 0.

@Keneo1

5 ай бұрын

@@krishnannarayanan8819it means 0 or -0

@morijin5568

5 ай бұрын

@@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.

I haven't been to school for at least 5 years, and now listening to your explanation I could understand better what I could not back then

never took a calculus class in my life but i still end up watching these videos

@jamescollier3

5 ай бұрын

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

@muneebmuhamed43

5 ай бұрын

studying in class 10 but still watched cuz why not 😂

@jim2376

5 ай бұрын

Admirable curiousity. 👍

@dominicj7977

5 ай бұрын

​@@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

@operator8014

5 ай бұрын

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

brilliant explanation, these videos really make me feel like I'm getting a better grasp on calculus as someone who's never taken it but is passionate about math :)

Are we not gonna talk about the amount of markers he's got stored in the background? I've seen entire schools have less than that lmao

Knowing how this function's graph behaves gives all of the intuition needed. Vertical asymptote at x=1, positive to the right, negative between 0 and 1. My students often dive into the calculus without thinking about the precalculus. Sure it can be done without the precalc, but the confidence gets a big boost when we think about the graph first.

@No-cg9kj

5 ай бұрын

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.

@bobtivnan

5 ай бұрын

@@No-cg9kj read more carefully

@bramvanduijn8086

5 ай бұрын

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.

@iamcoolkinda

5 ай бұрын

@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

@thetaomegatheta

5 ай бұрын

@@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.

Thank you. The limit pops up a lot in engineering, not just on paper but in actual physical or electrical functions. That said, in electronics we use graph paper as writing paper so doing a graph with 3 samples is quicker.

I love this bite sized math content, feels like I'm getting a bit smarter every day :D

I was taught that, if direct substitution results in A/B, where A and B are nonzero, that’s the limit. If you’re given a limit that is A/0, the limit DNE. If you’re given a limit that is 0/B, the limit is 0. If the limit is 0/0 it’s indeterminate. Methods like multiplying by conjugate, or L’Hopitals rule come in to play. So from first glance, you can instantly tell the limit DNE because the numerator is nonzero and the denominator is 0 when direct substitution is applied.

@lugia8888

25 күн бұрын

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

@wills4104

25 күн бұрын

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

@PixelVoyager777

18 күн бұрын

​@@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.

Finally! A problem on this channel I could solve on my own!

@RavenMobile

5 ай бұрын

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.

Btw, the graph shows that the two lines doesn't meet at a certain point (diverging on an asymptote) so that's what DNE literally means: there's no convergence at any particular point.

@bartiii7617

5 ай бұрын

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

@bartiii7617

5 ай бұрын

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

@hyperpsych6483

5 ай бұрын

@@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

@hydroarx

5 ай бұрын

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

@alexatg1820

5 ай бұрын

@@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

your teaching style is comforting i still dont understand this one due to my lack of foundational knowledge, i think. still very glad to have your vids

I just literally advance self-studying Calculus 1 rn and this is my 1st video yt recommend it. I didn't know that we can also have exponential signs to determine if + or - infinity but i alr knew that it will be DNE bcuz of + & - infinity are not equal. Perfect timing! I can't wait for my next sem. You got a sub❤!

@bprpcalculusbasics

5 ай бұрын

Glad to hear! Thank you!

@zxcvbn089

Ай бұрын

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

It's been a while since I did this type of stuff. Thanks for the refresher.

I haven't taken calculus in my life and was interested. Only to be completely distracted by the lifetime supply of expo markers.

Merry Christmas Bprp 🎉

@bprpcalculusbasics

5 ай бұрын

Thank you! You too!

I once drew a piecewise function for my learners, below 2 it was defined as x and from 2 and above it was defined as x+2, I asked them to determine the limit of the function as it approaches 2. I realized they had long forgotten about approaching a point from the left and right, they've gotten used to simplifying then "substituting" the limit value.

Merry Christmas Mr. bprp!

I always loved problems like this, it always reminded me that when u set up a number line the distance btw what ever numbers you end up choosing is infinite and if you wanted to count every number btw that distance you would always be approaching a certain number and never really reaching it.

Almost all limits can be evaluated by doing a thought experiment of "what happens if I move just ever so slightly to the left/right/both sides of it" - and then playing it out in your head.

in my opinion, this question is the best way to tell us why concepts are important.

We always learned to take the limit approaching form the left, the right and then combine them

I personally prefer using two sequences to show that. X_n=1+-1/n, and then the functions turns to to 1+-n, and when you take its limit you get +-infinity. Feels more rigorous to me.

Thankyou for revision 😀

Good video as always

Nice video! It’s been a while since I’ve done this. Since we did indeed get the conclusion that we would expect from inspection, could you give an example of a limit that looks DNE at a glance, but turns out not to be?

@literallyjustayoutubecomme1591

5 ай бұрын

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

@Steve_Stowers

5 ай бұрын

@@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).

@jackbrax7808

5 ай бұрын

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!

@Steve_Stowers

5 ай бұрын

@@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.

@jackbrax7808

5 ай бұрын

@@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.

I was not getting LHL = RHL so I knew the limit does not exist. Btw Merry Christmas bprp :D

It’s been 16 years since I’ve taken any calculus but knew right away it was undefined or does not exist

I really like this explanation since this explanation shows us that we mathematician did not do math recklessly according to the writing only, but according to the meaning of the limit hidden in the math problem.

I've found another method. We know that x/(x - 1) = ( (x - 1) + 1 )/(x - 1) = 1 + 1/(x - 1). So the limit equals 1 + lim_(x -> 1) { 1/(x - 1) }, or just 1 + lim_(x -> 0) { 1/x } which we know DNE.

@colinjava8447

5 ай бұрын

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.

@janskala22

5 ай бұрын

@@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".

@colinjava8447

5 ай бұрын

@@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.

@janskala22

5 ай бұрын

@@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.

@colinjava8447

5 ай бұрын

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

This question gets a bit easier if you rewrite x/(x-1) as (x-1+1)/(x-1), which is 1 - 1/(x-1), which is a little more obvious in how it behaves as x -> 1. Fairly simple explanation though.

Unless you consider using the Riemann Sphere (extended complex plane). The only problems with it are when attempting to divide zero itself by zero, but we aren't trying to do that here, so the identity checks for all numerators not equal to 0.

I honestly think that people usually forget, checking the limits right and left to just verify that the limit exist, usually what they do and what I did initially is I added one and subtracted one from the numerator and some split it in kind that it becomes one plus one by X minus one extends to one. So the whole thing tends to infinity which tends to zero so overall the limit would have been one but that points out to be incorrect.

From the times, when I was a student, I remember three different intinities: "+∞", "-∞" and "∞". So we explicitly used sign, if the infinity had one, and not used if that was "just the infinity", when sign is unknown (or does not matter). Is this the case nowadays? You are never using "+∞" notation, always omitting "+" sign...

@Mr.Not_Sure

5 ай бұрын

Same

@thetaomegatheta

5 ай бұрын

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.

While I do like the idea of the plus on the zero meaning a number arbitrarily close to zero, for problems like this I always think of it a “positive” zero. It’s functionally the same and gets the same answer, I just find it easier to understand, if you divide a positive by a positive, you get a positive. Is the zero positive or negative, not really, but if anyone is having trouble understanding this, try thinking of it this way.

@fioscotm

5 ай бұрын

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

@tomekk.1889

5 ай бұрын

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

-X approaches a vertical asymptope from the left, whereas +X approaches a vertical asymptope from the right. If -X doesn't equal +X the limit does not exist.

Very good! thank you.

Is this a well-defined limit? Calculus question on Reddit r/askmath kzread.info/dash/bejne/iZV6p9OngKrgc9I.html

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. I'm just now trying to keep up with the material as we're quite past that and even had a small test (which I failed, miserably) and I'm going to definitely retake it soon as thanks to you I understand everything perfectly, even though English is not my first language haha Lots of love from Poland! Cheers!

@thetaomegatheta

5 ай бұрын

'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.

took me like 3 seconds but my lack of trust in myself and curiosity kept me here lol

We were taught that the general limit for that would not exist because one side goes to infinity and the other goes to negative infinity. You would have to do a directional limit

i feel like the answer is either positive or negative infinity, but it's not defined from which side are we approaching x, limits tend to have positive and negative zeroes that gives 2 possible outcomes for 1 limit so it just doesn't work

I personally think it's easier to approach these problems by changing the limiting value to 0. That way it's obvious what's +ve and -ve. In this case, we could change the limit to lim e->0 (1 + e) / e by substituting x for 1 + e (e is supposed to be epsilon here). Then you can work out lim e->+0 and e->-0 and it's a little easier (IMO).

It's normal, there is no limit. Ask the good question: what is the + or the - limit? The problem is often the nut behind the whiteboard.

@marvinliraDE

5 ай бұрын

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.

@marvinliraDE

5 ай бұрын

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.

@OnurOzalp-personal

5 ай бұрын

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

@clmasse

5 ай бұрын

@@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.

@dielaughing73

5 ай бұрын

​@@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.

If you've done the x/x limit before, it's super easy--graph looks the same just shifted 1 unit to the right due to the "-1" in the denominator.

The right usage of infinity is a gulp of fresh air.

I guessed that it was undefined in 2 seconds, cause its essentially 1/x, and its a limit from both sides.

Nice, this is exactly how I did it. Can you post more proofs for common derivatives using the limit definition of (f(x+h) - f(x))/h? I think it could be fun to do a whole series on that. I tried the polynomial one for myself, and was able to confirm that (x^n)’, using binomial theorem and being left with just the second term was nx^(n-1). I tried getting the derivative of e^x = e^x, but I couldn’t pull it off tho, wasn’t sure how to bring out the h 😂

@Syndicalism

5 ай бұрын

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

@darcash1738

5 ай бұрын

@@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?

it is ±infinite depends on which side you take. - if approach from -inf to 1 + if approach inf to 1. You can easily check using scientific calculator. Type in the function and calculate wiith x= n±10^(-6)

@thetaomegatheta

5 ай бұрын

'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.

Do you think a teacher or other person who check your work will be fine with 1+/0+ argumentation? I think better would be substitute y = (x - 1) so, we need to calc (y + 1) / y with y -> 0 then (y + 1)/y = 1 + 1/y, thus for y > 0 it (y + 1) / y > 1 / y, but 1/y -> infinity but for y < 0 it doesn't work. so I think just use some fact lim (x + C) = C + lim (x). idk.

@dielaughing73

5 ай бұрын

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.

@AryanRaj-fz7dd

5 ай бұрын

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

@billmilligan7272

5 ай бұрын

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.

@kazedcat

5 ай бұрын

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

@anonymousf7byyj

5 ай бұрын

@@billmilligan7272thanks for your input Karen

I am absolutely certain that my Calculus teacher from 20ish years ago would have hated your 0+ notation…. She wanted derivative tests all the way. EDIT: To clarify, she would obviously have been fine with 0+ in the initial limit, but she wouldn't have liked 0+ as a result of partial computation. [Though it does seem intuitive as shown in this video.] She would have considered this an "invalid shortcut".

@tomctutor

5 ай бұрын

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! 😟

@bernhardbauer5301

5 ай бұрын

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.

I still remember us being shocked when the teacher wrote positive and negative zero. We were perplexed, bamboozled even. Until he explained why and how.

Out of the corner of my eye I thought this was loss

DNE since f has a VA at x = 1 and f is odd

@epikherolol8189

5 ай бұрын

Vertical asymptote?

@thetaomegatheta

5 ай бұрын

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.

@xinpingdonohoe3978

Ай бұрын

f is not odd at all.

Of course class couldn't figure it out if no solution exists 😂

@isaacbruner65

5 ай бұрын

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

@omp199

5 ай бұрын

​@@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.

@kazedcat

5 ай бұрын

​@@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.

@omp199

5 ай бұрын

@@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.

@kazedcat

5 ай бұрын

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

I don't know why this video got recommended to me. I would not have know the correct notation, but figuring out the answer is quite simple.

Got it correctly! Remembering stuff from years ago that I learned from bprp

Looks like it just tends to ∞

@isaacbruner65

5 ай бұрын

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

@lucaspanto9650

5 ай бұрын

@@isaacbruner65 🤓

@thetaomegatheta

5 ай бұрын

@@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.

@aarusharya5658

5 ай бұрын

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

Why L hopital rule is not used here

@OK-ei7io

5 ай бұрын

We don’t have an indeterminate form.

@YourNeighbourJack

5 ай бұрын

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

@Harishkumarindianrailways

5 ай бұрын

Understood

@teelo12000

5 ай бұрын

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

@J-M784

5 ай бұрын

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

Converges in the one point compactification of the reals 🤯

I think the issue is people are so used to the assumption that the limit exists so they just straight up applying any techbique before checking.

I figured out it DNE in 5s lol

@General12th

5 ай бұрын

Do you want a sticker?

@Sukunut

5 ай бұрын

@@General12th lol

my first thought it should be the same as lim (x->0)1/x (substract 1 and add 1in the numerator, simplify to lim(x->1) 1 + 1/(x-1), 1 is a constant and don't really change anything here and finally lim(x->1) 1/(x-1) looks almost like lim (x->0) 1/x, which explanation, usually always given in a class rooms).

Wow, I still remember calculus!

damn left and right limits really got me on this one

If the limit tends to infinity, it does not exist, because infinity is a concept, not a number. What infinity means in this context is that as we get closer to the limit, the value is always greater.

I did all this in my head. No joke. I know the number to the left of 1 minus 1 must be negative 0.00001 (something like that) and the number to the right of 1 minus 1 must be positive 0.00001 so I can include the right-handed limit approaches positive infinity and the left-handed limit approaches negative infinity, so the general limit does not exist. Easy peazy lemon squeezy.

You can use graphical methods too: x = x - 1 + 1 ==> x/(x - 1) = 1 + 1/(x - 1). So y = x/(x - 1) is basically the 1/x graph shifted to right by 1 and up by 1 unit. As x -> 1, x/(x - 1) diverges. So limit does not exist if you know the 1/x graph well.

I always imagine coming from the left and right to approach the number in such cases (also I make gestures with my fingers 👉🏻👈🏻😂)

OMG! What should I watch, your T-shirt or the math 😆

this is called sidded limits, when we know that theri is a discontinuity but the plot has one side going up and the other going down. The plot makes it easier to understand.

@xinpingdonohoe3978

Ай бұрын

Technically, we should check sided limits, and infimum and supremum limits, for every limit, but the continuity of some functions makes it fine normally.

This is incredible. I cannot believe it. This has got to be some kind of a record. A miracle. I tried it and I got it right. I never get things right. This got to be the exception to prove the rule. Wait...

You can also answer lim_x→1[x/(x-1)]=∞ without precising the sign, and it means that there is an asymptote at x=1 in this case, without telling from where the function approaches it. Be careful not to mistake it for lim=+∞, tho, this one would mean that the limit is defined. This is also why I would recommend to always say +∞ instead of just ∞ when the value is positive.

@xinpingdonohoe3978

Ай бұрын

Like square roots, common practice has people being sloppy. They don't do much to distinguish, for example, positive infinity from unsigned infinity, or the multivalued square root from a single branch (generally the principal one).

I graphed it and it approaches negative infinity from the left and positive infinity from the right so if I remember my calculus correctly that means the limit doesn't exist, although the one-sided limits do. Now I will watch the video and see if my 2-second-effort-because-I'm-not-in-college-anymore answer is correct... Yay I got the right answer on a math video for once!

REally cool. Anyways, Merry Christmas!!

@bprpcalculusbasics

5 ай бұрын

Thanks! Merry Christmas to you as well!

I could only figure this out if I drew the graph and realized that the positive and negative directions gave different answers. But I think I would have gotten this test question wrong because I would have just used simple substitution and came up with positive infinti.

any infinity limit DNE. +inf, -inf and inf are just handy concepts to describe certain cases. Sayin that some inf limit does exist because it has a sign and other does not is just word gymnastics

Damn I wasn’t taught that there are two answers. I just said positive infinity instantly without knowing that you have two answers , both negative and positive, and have to check left and right side to see which one is correct

At 5:43 the denominator should be written as -0(plus). It is a negative number, with a magnitude just larger than zero.

this is easy lol idk how the class didn't figure it out

great explanation!

I tried to substitute u=x-1 so I could get a simpler (in my head) expression of lim_u->0[(u+1)/u] = lim_u->0[u^-1 + 1], which I just know the graph of and can see that it's indeterminate

Also, even if both the 1+ and 1- limits became the same(∞ or -∞), the limit would still not exist because limit only exists when Left hand limit= right hand limit= finite quantity

This is fucing essential

I think I found my problem with high school math. Getting distracted. Such I was by this video by that impressive ready to go supply of dry erase markers in the lower right hand of the screen

I approached it graphically. Let X divided by (X-1) equal Y, then start graphing a few points. For very large values of X, Y approaches 1 (from above). You're dividing large numbers by a number just slightly smaller. As X gets closer to 1 (from above), the graph turns up and heads toward positive infinity. Turn that around and for very negative values of X, you're dividing X by a number just slightly more negative than X. So Y approaches 1 (from below). As X gets closer to 1 (from below), Y gets smaller. It crosses the origin at X=0, then goes negative and heads off to negative infinity. Been too long for me to recall if this is a hyperbola or something else. But it's still a fun little question, and an important light bulb to turn on in students.

Good old math yay! I simply said "negative infinity" since I only considered x approaching 1 as x = 0.99999999, but yeah it could have been x = 1.000000001 too.

everytime you have a limit question, divide both the numerator and denominator by x^n

@carultch

3 ай бұрын

That works if you are looking at the limit as x approaches infinity, or negative infinity. You just compare the highest ordered powers of x, and ignore all the other terms. If they are both the same power, the limit as x approaches either infinity is the ratio of coefficients. A higher power on top, means it approaches one of the two infinities, depending on the sign of the coefficient and the net power on the ratio of x. A higher power on bottom, means it approaches zero. For limits at finite values of x, it's the net multiplicity of poles and zeros at that point that matters. I.e. number of poles minus number of coinciding zeros. More poles than zeros means the limit doesn't exist; more zeros than poles, means the limit is zero.

My Teacher in the L1 of Limits told me about the 8 indeterminable forms: 0/0, inf/inf, inf - inf, 0^0, inf^0, inf*0, 1^inf. Anything except these including 1/0 [inf] is defined {though the limit may not exist}.

@xinpingdonohoe3978

Ай бұрын

That's 7, not 8. There are indeed 7. But it's a bit of a stretch to say anything else is "defined". It's just that everything else is "predictable". This 1/0 will go to some sort of ∞, but we must check how and which.

Thank you very much

Just a question, the given limit wouldn’t express the fact that the function, when approaching 1 from the negative side, goes to -infinity while approaching 1 prof positive side goes to +infinity?

@carultch

3 ай бұрын

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.

a quick graph on a spreadsheet is faster in many cases, to know whats going on

Getting heavy Mean Girl vibes here...

havent taken any difficult math classes…. yet. but i just put in 0.5, 0.75, etc and could tell that diving something getting closer to 0 is infinity, but its infinity?