Real Analysis | Algebraic Properties of Limits

[@drpkmath12345]

Well explained. More details than Rudin textbook for this topic I guess

[@dr.pkmath319]

Isnt it Rudin? Maybe you made a typo

[@drpkmath12345]

Nathan Scott haha yes Rudin! That was a typo thanks for pointing out~

[@dr.pkmath319]

MathFlix Not a problem bro that happens

[@froglet827]

Abott>Rudin

[@ghislainleonel7291]

Thanks for this insightful video on basic proofs of limt properties

[@curtisjuelz2327]

I know im asking randomly but does anyone know of a trick to log back into an instagram account..? I somehow lost the login password. I appreciate any tips you can give me.

[@rsmagl693]

At 17:35 how did |bn-B|＜|B|/2 imply that |bn| >|B|/2

[@beaumaths]

|bn-B| < |B|/2 means that the distance between bn and B is less than |B|/2. We can rewrite this as B-|B|/2 < bn < B+|B|/2. Basically bn is inside the interval (B-|B|/2, B+|B|/2). If B is positive, then B-|B|/2=B-B/2=B/2. Since that is the lower bound of the open interval, bn>B/2=|B|/2 If B is negative, then B+|B|/2=B-B/2=B/2. Since that is the upper bound of the open interval, bn

[@Claymaker808]

@@beaumaths a

[@alegal695]

Because of the inverse triangle inequality: | |x| - |y| |

[@elgourmetdotcom]

I love all these kind of proofs!! I would very much appreciate a complete lhopital rule proof, there are many non rigorous out there

[@jkid1134]

Epsilon delta stuff is so cool. Very nice video.

[@taopaille-paille4992]

a cool use of epsilon delta stuff are Cesaro-style theorems / lemmas.

[@goodplacetostop2973]

20:41

[@hyperboloidofonesheet1036]

I'm still having trouble with #4. I understand that the limit doesn't exist if B = 0, but if I define a sequence of b values where at least one of the values is zero, but B is nonzero (for example, b[n] = 10-n), then even in the case where a[n] = 1, you have an individual value that is undefined (in this case a[10]/b[10]). It seems the restriction on the b sequence is stronger than just its limit being nonzero -- it can't contain _any_ zero values.

[@tracyh5751]

Correct. a_n/b_n is not even defined if b_n=0 for any n.

[@tracyh5751]

Many people will also say @ゴゴ Joji Joestar ゴゴ is correct. Here is how the two views come together: If b[n] =0 for some n, the sequence a[n]/b[n] is not well defined, so some would say it doesn't make sense to even speak of a limit for this not even defined sequence. However, if the limit of b[n] as n approaches infinity exists and is nonzero, b[n] can only equal 0 finitely many times. That means that there is an N so that b[n] is not zero for all n>=N. We can then consider the sequence a[n+N]/b[n+N] which you can think of as the sequence a[n]/b[n] but we have removed the first N terms. We can then take the limit of this new sequence, and its limit will be equal to the limit of a[n] divided by the limit of b[n] (provided both of these limits exist and the limit of b[n] is nonzero).

[@hyperboloidofonesheet1036]

@@angelmendez-rivera351 This would sound correct, but I'm agreeing with Tracy H. You can't call a[n]/b[n] a sequence if for some values of n the result is undefined or infinite. If you allow infinity into the mix, then you break the rule than a sequence which converges is also bounded, and this unravels other parts of the presented proof.

[@proexcel123]

At 10:42, can we just choose |bn| = B since it converges to B, instead of using the bounded property and choose |bn| < M?

[@mohamedaminehadji6415]

No, we cannot because b_n is not equal to B for all n. Your way of reasoning is correct, but you have to say "b_n converges to B, so b_n is *close* to B, so |b_n| is bounded from above" (which is the bounded property)

[@proexcel123]

@@mohamedaminehadji6415 Thank you! Actl after one year as a Uni student, I kinda got that down alr. But thanks anyways!

[@benjaminbrady2385]

Why are some of the videos private in the Real Analysis playlist? Loved the video btw!

[@vikaskalsariya9425]

he records the videos in bunch and releases 1 per day. Almost every math channel does this.

[@tomatrix7525]

@@vikaskalsariya9425 yep, sometimes it is probably easier I’d guess to do sessions of recording in different topics where he would record like 4 or so videos.

[@honest-bear-1246]

thank you

[@ayoubnouni224]

In 11:04, i think you have to exclude the case when A=0

[@ayoubnouni224]

It’s okey you’ve mentioned it later 😅😅

[@fndTenorio]

5:40 I dont't get why |an - A| + |bn - B| < ε. Edit: Ok, now I see he is working in reverse.

[@ammjjackson821]

Thankhs Pro

[@chriskojulesvidal7434]

CAN YOU EXPLAIN HOW YOU SEE VERY EASILY THAT |Bn-Ｂ｜＜Ｂ/2 IMPLIES THAT |Bn| >|B|/2

[@davidmoss9926]

Draw a picture. If the |bn| is within |B|/2 of |B| you can observe that the |bn| must be greater than |B|/2

[@medwards1086]

By the triangle difference inequality we have ||b_n|-|B||

[@johnstroughair2816]

For the last result why not just use the fact that bn is bounded

[@alegal695]

How would you use it? Share your proof and we can discuss.

[@dancetoyourownrhythm5975]

Yo 🤪from India.... Thanx sir