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!
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{...}
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(:
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.
3 ปีที่แล้ว +1
That's a neat (probably the most neat) explanation on this type of limit. Can't thank you enough
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.
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|)
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 :)
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 !!!
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.
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 "
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
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
@@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.
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!
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.
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
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
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.
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
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 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
So if epsilon is let's say 50, if I just leave delta fixed at 1 then it's always going to below epsilon because epsilon is so large. It'll work equally for epsilon 500 or 5 million too. But as epsilon gets small we need a more careful choice of delta as it's easy for a bad choice to be above epsilon. In this smaller regime delta being epsilon over 5 is a good choice. Of course we could choose delta=e/15000 and it'd work too but we use algebra to derive the maximum delta that works as it's easier to prove that value than some smaller weirdly selected value.
great video but fell flat on the proof in the last 3 minutes. No understanding of what you are getting at after 11:00, although I do get the components from earlier in the video
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.
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?
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.
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.
You can pick any δ you want. The only thing will the constant 5 in the expression ε/5. It will be smaller if you pick a smaller δ and larger for a larger δ.
If you send me your proof I can check it, but it's hard to tell by just looking at a value of delta. What you need to show is that if |x-2|< (your delta) then |x^2-4| < epsilon
kushbu r Yes, this method is fine! The 'hard' part of this problem is finding a bound on |x+2|, we do it one way in the video, you are finding a bound by using the triangle inequality. Just something to note, what you've done is "scrap" work to find delta, to do the actually proof you would want to do something like Let d = -2 + sqrt{4 + e}. Then |x^2-4| = |x-2||x+2| < (d+4)d (by triangle inequality) = d^2 + 4d = e Which shows that as long as |x-2|< d then |x^2-4|< e
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.
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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?
This is the best epsilon-delta explanation I've ever seen !
Thank you !
The best explanation on THE WHOLE INTERNET!!!
After watching countless videos, this is the first one that makes some sense. Thanks.
These videos are 11 years old and really helping me now
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.
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(:
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
This is the clearest explanation of the epsilon-delta proof i've ever seen. so amazing.
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
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.
That's a neat (probably the most neat) explanation on this type of limit. Can't thank you enough
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.
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.
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.
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.
God bless your soul I have fallen in love with this video, only one that I have found that makes logical, sequential sense!
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|)
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 :)
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
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 💕💕💕
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.
this is really brilliant way to think about solving this type of problem.
i love you so much. You've cleared up much confusion.
Thank you very much for this great video. I finally understand how to use the epsilon-delta definition of a limit to prove limits. :)
This is the best explanation I saw.
Thank you!
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.
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 "
nice video man. really helping me out
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.
a question, how can we know that C|x-2|< Epsilon?
Man, it would had been wonderful to have a classmate like you in school
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.
Why |x-2|5 < epsilon???. I've already watched many videos of this definition and i still don't understand this part
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
Does anyone know what kind of tools were used for writing in this video? It is great.
absolutely the best explanation
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
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
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!
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.
sir can i take delta be equal to epsilon over absolute value of x+2?
Very clear explanation. Great work.....would be nice if you could make a video on the lim sinx/x .
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
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
I was having so much trouble with this, thanks for making this video. I understand now
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!
Why do you choose that delta is 1? What if I chooses that delta is 1.5 for instance? Would I get wrong answer?
|x-2|
Awesome video, with it I could get along with my calculus course.
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
Thank you so much for the clear explanation! Helped me a lot with my homework!
@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.
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
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
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.
what if its X-> (-2)? could you do the same problem but approaching a negative number?
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"? :)
Great video with great explanation it help me lot, so thanks and keep uploading.
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
So if epsilon is let's say 50, if I just leave delta fixed at 1 then it's always going to below epsilon because epsilon is so large. It'll work equally for epsilon 500 or 5 million too.
But as epsilon gets small we need a more careful choice of delta as it's easy for a bad choice to be above epsilon. In this smaller regime delta being epsilon over 5 is a good choice.
Of course we could choose delta=e/15000 and it'd work too but we use algebra to derive the maximum delta that works as it's easier to prove that value than some smaller weirdly selected value.
I do have one! Search on youtube for "rootmath sinx/x" and you should find it
so more formally, it is a limit problem within a limit problem?
Where does the 5 come from?
HUGE Thanks! I finally get to know this!!!
great video but fell flat on the proof in the last 3 minutes. No understanding of what
you are getting at after 11:00, although I do get the components from earlier
in the video
9:49, "delta"?
Why we choose x=3 rather than x=2?
10/10 explanation. BRAVO!!!!!
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.
After watching so many videos on this ,what l am not getting what are we trying to do..😂
Me too 😂😂
Thanks, very well explained, excellent video.
You saved my life
Nice explanation. Tranks! :)
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
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?
sir, do you do that with mouse or those table pen for computers that idk the name xd
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
God bless this video, God bless this man, God bless that cute neat little graph I think I finally get it now
Really really clear. Thanks
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.
Awesome explanation.
can |2+x|
Gbat Noche sure, if 1 < x < 3 then 3 < 2 + x < 5. Thus 2 + x is positive so |2 + x| = 2+x < 5
|2+x|
Not sure exactly why, but, in short, it is safe to say | x + 2 | < 5 (which means -5 < x + 2 < 5).
Doesn't this only work for ∂ ≤ 1?
You can pick any δ you want. The only thing will the constant 5 in the expression ε/5. It will be smaller if you pick a smaller δ and larger for a larger δ.
finally i have found a good proof.
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 did this differently and got delta=-2+root(4+epsilon)...is this also right, if not can someone explain why not?
If you send me your proof I can check it, but it's hard to tell by just looking at a value of delta. What you need to show is that if |x-2|< (your delta) then |x^2-4| < epsilon
rootmath Thanks! sorry can’t find epsilon but I’ve used e and d for delta and
kushbu r Yes, this method is fine! The 'hard' part of this problem is finding a bound on |x+2|, we do it one way in the video, you are finding a bound by using the triangle inequality. Just something to note, what you've done is "scrap" work to find delta, to do the actually proof you would want to do something like
Let d = -2 + sqrt{4 + e}. Then
|x^2-4| = |x-2||x+2| < (d+4)d (by triangle inequality) = d^2 + 4d = e
Which shows that as long as |x-2|< d then |x^2-4|< e
rootmath okey doke, thanks a lot! :)
I understand everything except the part where you say d = -2+root(4+e) how did you conclude that with only knowing that d^2+4d=e
very good explanation
Very well done!
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.
Have been a big help!
Great explanation. Thanks.
Obrigada pela explicação! saudações em pt BR 😊👍
All you gotta do is plug in 2 into the equation
Why can’t this work with algebra?
Thanks awesome video!