The choice of 1 is completely arbitrary, you can pick any number you want. The bound on |x+2| will change depending on what number you pick, but that's not a problem.
Hey @rootmath, can we get some more epsilon delta proofs. I believe to have through all your videos on epsilon delta limits [3] and including this one 2 proves.
"That may have been more explanation than you wanted" (time 8:02) Ha! Let me assure you: it is better to have too much explanation than no enough. This same example was in my Calculus book, and I've been racking my brains as to how to understand it. This video helped me make sense of it. In retrospect my Calculus book did hit all the same points you did but far, far less thoroughly. So thanks, Rootmath, for your thorough explanation!
oh my god.Bless your soul!!! I have been struggling with this whole concept for SO long and you just clarified EVERYTHING. I should be giving all my tuition to you instead of a school where all the instructors fail to teach anything. Like omg I am so grateful!!! you couldn't have explained it better!!
At first I thought this was easy, but now it's probably my 9th time watching this video and I realized or I'm really dumb or this shit is some kind of hard and confusing
It's really hard. Although I can understand 3 Calculus' content, I can't understand on the whole what this definition and their proofs mean. I expect we (or at least I) can get it. understand it
Well yep i ve been replaying several different videos as of now, keep coming back to khan ones, or some other channel, and also my college's channel, since yesterday. I almost intuitively understand it.. if not for the proving part after getting ∆=min{...}
This is an example in my textbook, and the solution was only few lines!!!!! I am not wondering anymore why I didn't understand it from the textbook. This was so much better than my textbook. thanks
dont watch this video it is bad at explaining look for other ones i struggled way to much here because there are lot of steps that has not been done but in other videos it is muc hsimpler
Finally, an explanation of the comparison example that is extensible to cases like the dreaded (x^2+2x+4)-12 that comes up in every freakin analysis book.
I know this is a long time after the video was made but this was an incredible explanation of the delta-epsilon definition along with the precise definition of limits. Thank you so much for your help!
We basically decided to let delta always be less than one. In that case, no matter what the value of x is in our range, (x-2) will ALWAYS be smaller than 5. Looking at the graph, you can see that is proven to be true. You then have proved that x+2 will always be less than 5, so you can then substitute 5 in for x+2 knowing that 5 is a "worst case scenario" for how big x+2 can be.
At 8:39, why isn't it less than or equal to ε? Since |x-2||x+2| is essentially |f(x) - L | and since |x-2||x+2| < |x-2|·5, can't ε be greater than or equal to |x-2|·5? Any clarifications are appreciated(:
it will still be the same result, just that, as rootmath said, the value of |x+2| will then be a different value (you used 1.5 as an example, in which case |x+2| would just be 5.5) but the way that you would proceed to do the proof would be exactly the same :)
A more concrete example of min() , and max() would be: min(a,b)=(1/2)*(a+b - |a-b|) and max(a,b)=(1/2)*(a+b + |a-b|). So threfore δ=(1/2)*(1+ ε/5 - |1 - ε/5|)
You made it sound like it was going to be very hard, but it wasn’t, thanks to your explanations. I had to pause a couple times, but there were no showstoppers. With regard to the chosen problem, I feel that an example that has no limit is just as important as one that does. (Anything besides lim x->0 of |x|/x because I feel it’s slightly unrealisic). Liked!
Thank you so much for clarifying! I feel like I've been cheated by my calculus professors when I took all three series (we didn't go through any of the proofs). This makes the topics in my analysis class much clearer. Now I understand why we pick the minimum of the two; it was never explicitly shown to me in that way.
Thanks! It really helped with my understanding of more difficult proofs involving limits. I agree with the replies to your comment about giving more explanation than someone needs. In my case, the extra explanation really helped understand the relationship that you wanted to convey. So, thanks for the extra mile.
Honest to God this is the best explanation out there thank you so much for all your effort. I Just subscribed and plan on binge watching all your videos
It’s not that you’re dumb, you just haven’t built your intuition enough to fully process this!!!! Keep practicing, you’ll get it!!!! I believe in you !!!
You're really good at explaining this stuff! Thank you so much! I feel more or less ready for a question like this if it appears on my calc midterm exam tomorrow! :D
You don't necessarily have to pick one, it's just an easy number to use that is close to x. You could choose a smaller Delta but then u be dealing with decimals, in the end you would still end up with epsilon
I'm pretty sure it's because if set delta = 1 then you get absolute value of 3 +2 (look at the graph) times the absolute value of x -2, but from that point you'll be shrinking delta so you know that it must be bigger
I feel like I missed something. If we are just making delta = 1 because epsilon is too big then how are we plugging in the value of epsilon/5 in for |x-2|? It seems like then we have to plug in 1 and then we would get 1*5=5 and get nothing to do with epsilon
First did you just pick delta to be equals to 1 and also delta can be 0.999 etc .explain why you picked delta to be equals to be 1 because there are alot of numbers that can be close to 2.
I could be wrong here, but at ~5:50, shouldn't delta be strictly less than 1 and not equal to 1? If delta = 1, than the interval surrounding 2 would INCLUDE 3. This would make the inequality |x + 2| < 5 incorrect. Because |3 + 2| < 5; |5| < 5; 5 < 5; which is obviously incorrect. I'm just curious if I'm missing something here? Thanks for any responses.
rootmath Thanks for the reply! Figured I was missing something. Well, not missing, just thinking incorrectly. Actually, could we pick almost any number for delta? Such as if we picked a delta = 2, the final delta would simplify to (epsilon/6) which still solves the proof, right?
pleasedie1979 Sorry this is so late, but yes, the choice of delta=1 is completely arbitrary, you can pick any positive number you want and the proof still works, it just changes the bound on |x+2| which also changes the denominator on epsilon as you pointed out, but the proof works the same way, just different numbers.
@@Lukinhas2012lk Basically, when delta = 1, |x - 2| < 1, so the values of x lies between 1 and 3. Therefore, |x+2| is < 5. Next, go back to |x - 2||x + 2| < epsilon, and plug in |x + 2| as 5, so it follows that |x - 2||x + 2| < |x - 2| * 5. So, this reduces the problem such that IF |x - 2| * 5 < epsilon, THEN |x + 2| * |x - 2| will also be less than epsilon.
It comes down to trial and error. He choose 1 (I presume) because it was an easy number to try and work with. For example he could've chosen the number 2 to be delta and and that would mean his corresponding x value would've been 4, and if you plug 4 into the function he would've gotten epsilon/6. He was just trying to find an upper bound for (x+2) so he could convert it into a constant value. So long as you stated that delta had to be less than or equal to 2, I believe choosing 2 as a delta value would've have also worked.
Thx for your great videos... I have question : on what bases do we assume that delta =1?? , maybe our assumption is wrong and maybe delta max is lets say 0.8 in other example with different f(x)?? ... i mean its not always guranteed that delta=1 is a correct assumption for all functios!. Thank you in advance
Thanks for the comment. delta=1 is completely arbitrary, you can choose any number, the bound we get will change but it's ok bc the problem will work out really similarly
Probably true for some, but sympathizing with people who already feel it's difficult may help relieve some stress and feelings like "if this is easy for everyone else then why don't I get it?". I think the main point is, this tends to be an area many struggle in, so if you're struggling that's normal.
Markus Our restrictions on x are based on the value of δ. The bigger δ is, the more values we can choose for x, so we don't want to make δ dependent on the value of x, because then each variable would actually be dependent on the other for its value, which makes things really weird. If δ had a large value, then by the equation δ=ε/|x+2| the x term would have a small value, and since δ is the possible range for all x values, requiring x to have a small value would also make δ small, consequentially making x large, and therefore necessitating a large δ and the process repeats. Putting an x term in the δ definition results in a paradoxical situation that we all want to avoid. It's like telling a computer to "ignore this command", the result is really bad.
Incorrect question. In order for this limit (the one we are trying to prove or maybe disprove) to exist we need |(x-2)(x+2)| be less than any epsilon when 0
Remember what we are trying to prove here. We need to find the ε greater than |f(x) - L|. Which looks like this: |f(x) - L| < ε Our function is x^2 and our limit = 4, so we get; |x^2 - 4| < ε we factor the above and get |x - 2||x +2|. In the video, this is where he wrote the "< |x - 2|*5 he can write the "
One question, why can't we choose something greater than 1, say, 1.5? I mean, the choice of 1 in here seems somewhat arbitrary. Could you clarify a little bit on that please?
I’m this case you can actually use anything bigger or less than 1, granted that it’s positive, it’ll just change the bound you put on |x+2|. There are some cases where you have to watch out for what you choose to take as delta as there could be asymptotes that cause problems.
All we care about is getting the left side to be smaller than epislon. We know that delta is less than epsilon/5 since it is the minimum. So we can use the epislon\5 we can also use the 1, but that won't be helpful, its not what we are trying to show.
The choice of 1 is completely arbitrary, you can pick any number you want. The bound on |x+2| will change depending on what number you pick, but that's not a problem.
Hey @rootmath, can we get some more epsilon delta proofs. I believe to have through all your videos on epsilon delta limits [3] and including this one 2 proves.
@@josuearreaga7142 yeah, I will try to make some more soon! Good idea!
@@rootmath you are the best! I'll be waiting^
Thankyou so much!!
Thanks, you are the first one, who explained me properly👌
"That may have been more explanation than you wanted" (time 8:02)
Ha! Let me assure you: it is better to have too much explanation than no enough. This same example was in my Calculus book, and I've been racking my brains as to how to understand it. This video helped me make sense of it. In retrospect my Calculus book did hit all the same points you did but far, far less thoroughly. So thanks, Rootmath, for your thorough explanation!
What software and table are you using for this video?
oh my god.Bless your soul!!! I have been struggling with this whole concept for SO long and you just clarified EVERYTHING. I should be giving all my tuition to you instead of a school where all the instructors fail to teach anything. Like omg I am so grateful!!! you couldn't have explained it better!!
The best explanation on THE WHOLE INTERNET!!!
This is the best epsilon-delta explanation I've ever seen !
Thank you !
At first I thought this was easy, but now it's probably my 9th time watching this video and I realized or I'm really dumb or this shit is some kind of hard and confusing
It's really hard. Although I can understand 3 Calculus' content, I can't understand on the whole what this definition and their proofs mean. I expect we (or at least I) can get it. understand it
Well yep i ve been replaying several different videos as of now, keep coming back to khan ones, or some other channel, and also my college's channel, since yesterday. I almost intuitively understand it.. if not for the proving part after getting ∆=min{...}
Oh mann i finally understand it.. 12:40 really clarified things
I can relate but hang in there. But I have come again all the way from learning real analysis to actually understand why we are doing this.
After watching countless videos, this is the first one that makes some sense. Thanks.
This is an example in my textbook, and the solution was only few lines!!!!! I am not wondering anymore why I didn't understand it from the textbook. This was so much better than my textbook. thanks
I thinks it's the 5th time in 2 years rewatching this
dont watch this video it is bad at explaining
look for other ones
i struggled way to much here because there are lot of steps that has not been done but in other videos it is muc hsimpler
These videos are 11 years old and really helping me now
That's a neat (probably the most neat) explanation on this type of limit. Can't thank you enough
This is the clearest explanation of the epsilon-delta proof i've ever seen. so amazing.
Finally, an explanation of the comparison example that is extensible to cases like the dreaded (x^2+2x+4)-12 that comes up in every freakin analysis book.
I know this is a long time after the video was made but this was an incredible explanation of the delta-epsilon definition along with the precise definition of limits. Thank you so much for your help!
Why are you such a legend in logic and explanation
Best video on epsilon-delta proofs! Cleared up all my doubts regarding the concept, thanks.
God bless your soul I have fallen in love with this video, only one that I have found that makes logical, sequential sense!
Very helpful - thanks for the clear description. This is the kind of thing my maths book whizzes through without telling me the reasoning behind it.
Nursultan Sulaymanov We are just graphing x+2 so that we can see that when x gets close to 2, x+2 doesn't get too big.
at 8:42, why can we say that
|x-2|5
We basically decided to let delta always be less than one. In that case, no matter what the value of x is in our range, (x-2) will ALWAYS be smaller than 5. Looking at the graph, you can see that is proven to be true. You then have proved that x+2 will always be less than 5, so you can then substitute 5 in for x+2 knowing that 5 is a "worst case scenario" for how big x+2 can be.
At 8:39, why isn't it less than or equal to ε? Since |x-2||x+2| is essentially |f(x) - L | and since |x-2||x+2| < |x-2|·5, can't ε be greater than or equal to |x-2|·5? Any clarifications are appreciated(:
it will still be the same result, just that, as rootmath said, the value of |x+2| will then be a different value (you used 1.5 as an example, in which case |x+2| would just be 5.5) but the way that you would proceed to do the proof would be exactly the same :)
Thank you very much for this great video. I finally understand how to use the epsilon-delta definition of a limit to prove limits. :)
I like how you pointed out how |x-2| being very small may make |x+2| very big. That intuition is missing in most of the proofs like this.
Wow thank you for explaining delta is 1 or epsilon/5. Alot of people skip that. Thank you.
Why can't our teacher teach us like you 😓😓 thank you sooo much 💕💕💕
Man, it would had been wonderful to have a classmate like you in school
A more concrete example of min() , and max() would be: min(a,b)=(1/2)*(a+b - |a-b|) and max(a,b)=(1/2)*(a+b + |a-b|). So threfore δ=(1/2)*(1+ ε/5 - |1 - ε/5|)
Thank you very much i found this video really helpful. They gave almost no explanation in the textbook and skipped right to the epsilon/5 part.
You made it sound like it was going to be very hard, but it wasn’t, thanks to your explanations. I had to pause a couple times, but there were no showstoppers. With regard to the chosen problem, I feel that an example that has no limit is just as important as one that does. (Anything besides lim x->0 of |x|/x because I feel it’s slightly unrealisic). Liked!
my math teacher is the best math teacher ever!
she taught us that we don't have to say ln , we coud pronounce it as lin.
#bestteacherever
Thank you so much for clarifying! I feel like I've been cheated by my calculus professors when I took all three series (we didn't go through any of the proofs).
This makes the topics in my analysis class much clearer. Now I understand why we pick the minimum of the two; it was never explicitly shown to me in that way.
Thanks! It really helped with my understanding of more difficult proofs involving limits. I agree with the replies to your comment about giving more explanation than someone needs. In my case, the extra explanation really helped understand the relationship that you wanted to convey. So, thanks for the extra mile.
this is really brilliant way to think about solving this type of problem.
i love you so much. You've cleared up much confusion.
Honest to God this is the best explanation out there thank you so much for all your effort. I Just subscribed and plan on binge watching all your videos
absolutely the best explanation
This is the best explanation I saw.
Thank you!
I am far too dumb for this.
Ikr!
It just takes time, practice, and repetition. Eventually you understand some of it.
@@juanlinde9028 only some?
It’s not that you’re dumb, you just haven’t built your intuition enough to fully process this!!!! Keep practicing, you’ll get it!!!! I believe in you !!!
Its not you, its everyone when they first encounter this. Just hang in there you will get there eventually
Why |x-2|5 < epsilon???. I've already watched many videos of this definition and i still don't understand this part
I was having so much trouble with this, thanks for making this video. I understand now
yoh thanks a lot, i was stranded with an example just like the one you did. I appreciate this a lot. please Keep doing what you do.
nice video man. really helping me out
a question, how can we know that C|x-2|< Epsilon?
You're really good at explaining this stuff! Thank you so much! I feel more or less ready for a question like this if it appears on my calc midterm exam tomorrow! :D
How do you know it is 1 and not 20 next to the 2? What if i choose 3?
always pick 1
+Garrett Roberts that was 11 months old, but thanks
You don't necessarily have to pick one, it's just an easy number to use that is close to x. You could choose a smaller Delta but then u be dealing with decimals, in the end you would still end up with epsilon
So, how do you know that |x-2|5
Same doubt, reply if anyone discovers
It was assumed. It's as in proving P -> Q, we assume P and show that Q.
I'm pretty sure it's because if set delta = 1 then you get absolute value of 3 +2 (look at the graph) times the absolute value of x -2, but from that point you'll be shrinking delta so you know that it must be bigger
so that we can choose delta = epsilon/5. that's the rationale behind it
we don't know that |x-2|5
Awesome video, with it I could get along with my calculus course.
Very clear explanation. Great work.....would be nice if you could make a video on the lim sinx/x .
Thank you so much for the clear explanation! Helped me a lot with my homework!
I feel like I missed something. If we are just making delta = 1 because epsilon is too big then how are we plugging in the value of epsilon/5 in for |x-2|? It seems like then we have to plug in 1 and then we would get 1*5=5 and get nothing to do with epsilon
After watching so many videos on this ,what l am not getting what are we trying to do..😂
Me too 😂😂
Really really clear. Thanks
Great video with great explanation it help me lot, so thanks and keep uploading.
First did you just pick delta to be equals to 1 and also delta can be 0.999 etc .explain why you picked delta to be equals to be 1 because there are alot of numbers that can be close to 2.
It doesn't matter what you pick delta to be, so just pick one where the arithmetic is the easiest.
Thanks, very well explained, excellent video.
10/10 explanation. BRAVO!!!!!
God bless this video, God bless this man, God bless that cute neat little graph I think I finally get it now
I could be wrong here, but at ~5:50, shouldn't delta be strictly less than 1 and not equal to 1? If delta = 1, than the interval surrounding 2 would INCLUDE 3. This would make the inequality |x + 2| < 5 incorrect.
Because |3 + 2| < 5; |5| < 5; 5 < 5; which is obviously incorrect.
I'm just curious if I'm missing something here? Thanks for any responses.
The video is correct. If delta = 1 then we see that |x-2| < 1, so then x < 3
rootmath Thanks for the reply! Figured I was missing something. Well, not missing, just thinking incorrectly. Actually, could we pick almost any number for delta? Such as if we picked a delta = 2, the final delta would simplify to (epsilon/6) which still solves the proof, right?
pleasedie1979 Sorry this is so late, but yes, the choice of delta=1 is completely arbitrary, you can pick any positive number you want and the proof still works, it just changes the bound on |x+2| which also changes the denominator on epsilon as you pointed out, but the proof works the same way, just different numbers.
Amazing explanation, but if you write that number "2" more like a number 2 instead of an "a" or "d" letter it would be less confusing
I really don't understand why we can't just take the inverse of x^2 apply that function to x+e and x-e and chose the minimum of the resulting output
HUGE Thanks! I finally get to know this!!!
at 8:41 , why is |x--2|5 < epsilon ? +rootmath
@@aiyopasta Because |x+2| < 5
@@aiyopasta Could you explain? I'm really lost
@@Lukinhas2012lk Basically, when delta = 1, |x - 2| < 1, so the values of x lies between 1 and 3. Therefore, |x+2| is < 5. Next, go back to |x - 2||x + 2| < epsilon, and plug in |x + 2| as 5, so it follows that |x - 2||x + 2| < |x - 2| * 5. So, this reduces the problem such that IF |x - 2| * 5 < epsilon, THEN |x + 2| * |x - 2| will also be less than epsilon.
@@aaronmei1630 why did we take delta less than or equal to one why not any other no.
@@snehahajong9288 you can use any number. 1 is just convenient
I do have one! Search on youtube for "rootmath sinx/x" and you should find it
Hello prof! Could you solve it for me please using delta epsilon proof: lim x approaches 9+ and then 9- of (9-x)^1/4=0 Thank you!
This didn't make sense at all. Why did you pick 1 and not 2
Have you found why one year later?
I honestly still don't know lmao! I totally forgot I wrote this comment haha
Wow! Those are A LOT of numbers!
It comes down to trial and error. He choose 1 (I presume) because it was an easy number to try and work with. For example he could've chosen the number 2 to be delta and and that would mean his corresponding x value would've been 4, and if you plug 4 into the function he would've gotten epsilon/6. He was just trying to find an upper bound for (x+2) so he could convert it into a constant value. So long as you stated that delta had to be less than or equal to 2, I believe choosing 2 as a delta value would've have also worked.
Are you guys serious? It is because of the definition of the limit, x is approaching 2 but it doesn't reach it. It means x/= 2
thanks a lot.
Thx for your great videos...
I have question : on what bases do we assume that delta =1?? , maybe our assumption is wrong and maybe delta max is lets say 0.8 in other example with different f(x)?? ...
i mean its not always guranteed that delta=1 is a correct assumption for all functios!.
Thank you in advance
Thanks for the comment. delta=1 is completely arbitrary, you can choose any number, the bound we get will change but it's ok bc the problem will work out really similarly
You saved my life
sir can i take delta be equal to epsilon over absolute value of x+2?
Why we choose x=3 rather than x=2?
Awesome explanation.
Why do you choose that delta is 1? What if I chooses that delta is 1.5 for instance? Would I get wrong answer?
Does anyone know what kind of tools were used for writing in this video? It is great.
Great explanation. Thanks.
very good explanation
Have been a big help!
Hypothesis: Telling people something is hard makes it feel harder than it would have otherwise
Probably true for some, but sympathizing with people who already feel it's difficult may help relieve some stress and feelings like "if this is easy for everyone else then why don't I get it?". I think the main point is, this tends to be an area many struggle in, so if you're struggling that's normal.
Very well done!
Hi! Can't you just say if |x - 2| |x + 2| < ε, then |x - 2| < ε/| x + 2| and define δ = ε/| x + 2|? Why would that not work?
Because we want delta to depend only on epsilon. The delta you have there depends both on epsilon and on x
Jorge Medina But we have restrictions on x, right?
Markus Our restrictions on x are based on the value of δ. The bigger δ is, the more values we can choose for x, so we don't want to make δ dependent on the value of x, because then each variable would actually be dependent on the other for its value, which makes things really weird. If δ had a large value, then by the equation δ=ε/|x+2| the x term would have a small value, and since δ is the possible range for all x values, requiring x to have a small value would also make δ small, consequentially making x large, and therefore necessitating a large δ and the process repeats. Putting an x term in the δ definition results in a paradoxical situation that we all want to avoid. It's like telling a computer to "ignore this command", the result is really bad.
why is x-2*5 less than epsilon?
Incorrect question. In order for this limit (the one we are trying to prove or maybe disprove) to exist we need |(x-2)(x+2)| be less than any epsilon when 0
so more formally, it is a limit problem within a limit problem?
The most people are focused on the proof.. is not the proof. it's the idea.. understand first the idea..than the proof is simple..
I don't understand at 12:00 where the < comes from, the one before Ix-2I 5. How does it relate to Ix^2 - 4I
Remember what we are trying to prove here. We need to find the ε greater than |f(x) - L|. Which looks like this: |f(x) - L| < ε
Our function is x^2 and our limit = 4, so we get;
|x^2 - 4| < ε
we factor the above and get |x - 2||x +2|.
In the video, this is where he wrote the "< |x - 2|*5
he can write the "
what if its X-> (-2)? could you do the same problem but approaching a negative number?
Obrigada pela explicação! saudações em pt BR 😊👍
Nice explanation. Tranks! :)
One question, why can't we choose something greater than 1, say, 1.5? I mean, the choice of 1 in here seems somewhat arbitrary. Could you clarify a little bit on that please?
@rootmath how did you choose the value 1 for delta ? can I use any bigger or less value ?
I’m this case you can actually use anything bigger or less than 1, granted that it’s positive, it’ll just change the bound you put on |x+2|. There are some cases where you have to watch out for what you choose to take as delta as there could be asymptotes that cause problems.
good video on hard problem. at 13:01, we want to pick the min of (epsilon/5, 1). you pick epsilon/5, but why not pick 1?
ok, you explained it--thanks
Hi, I'm spanish, I'm learning english. Can you tell me why you used "--" to conect "it and thanks"? :)
finally i have found a good proof.
Where does the 5 come from?
My only problem through the proof is how do you know the E/5 is the minimum of 1 & E/5? Could you please explain?
Todd Miller we don't know that E/5 is the minimum of 1 and E/5. We DO know that delta = min{1, E/5} so certainly delta
thank you sir
|x+2|
Nice proof!
Thanks. Great video.
why in min13 he took delta for {x-2} which is equal epsilon\5 and he did not take delta equals 1 for {x-2}?????????????
rootmath
All we care about is getting the left side to be smaller than epislon. We know that delta is less than epsilon/5 since it is the minimum. So we can use the epislon\5 we can also use the 1, but that won't be helpful, its not what we are trying to show.
sir, do you do that with mouse or those table pen for computers that idk the name xd
Awesome!