6 methods of evaluating the limit of a multivariable function (calculus 3)

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bprp calculus basics

Күн бұрын

Пікірлер: 38

@bprpcalculusbasics 2 ай бұрын

How to show nonexistence: kzbin.info/www/bejne/r3izpKKcicmnh9ksi=iPH8WT3GM-nsgFcA

@NixelKnight 3 ай бұрын

Thanks you so much! I don't have to take Calc 3 at my university, but I've always wanted to learn it. Your videos have been so helpful for me and I really appreciate it!

@andyloescher2401 3 ай бұрын

Take it for fun

@IisChas Ай бұрын

I laughed out loud at 15:30 when you drew the frowny face saying that there was no conclusion. I don’t know what it was, but it just got me.

@fireballman31 3 ай бұрын

The polar blooper lol

@Vernonnotyourman 3 ай бұрын

forgot to edit the middle part😂😂

@SuryaBudimansyah 2 ай бұрын

@@Vernonnotyourman It's basically becoming a feature in his channels

@bprpcalculusbasics 3 ай бұрын

Be careful when we use the polar method: kzbin.info/www/bejne/nX2Vl617gat5rbM

@leonardobarrera2816 3 ай бұрын

Thank for not editing that I appreciate that (Not sarcastic)

@eduardomarcicnetomarcic3511 3 ай бұрын

Thank you very much! Wonderful

@darth159 2 ай бұрын

Is there any way to tell whether you should start using the path method to show the limit does not exist or start the Squeeze Theorem?

@Gabes1321 2 ай бұрын

At 18:08 why can’t you just substitute r=0 into the bottom then leave it as (rcos^3@)/cos^2@ then cancel out the cos to get rcos@ then substitute r=0 again to get the limit as 0?

@krithikkumar9887 2 ай бұрын

Again what if theta is pi/2, then u can't cancel cos

@Gabes1321 2 ай бұрын

@@krithikkumar9887 two things. Doesn’t cos^2(@)/cos^2(@) approach 1 as @ approaches pi/2 anyway? Also as @ is an introduced variable, I thought we were allowed to cancel?

@thepotato1232 3 ай бұрын

Can someone explain me why (e^2x - 1)/x is equal to f'(0)? ( I only know math up to derivatives)

@t.b.4923 3 ай бұрын

one definition of derivatives is lim x->x_0 (f(x)-f(x_0))/(x-x_0). Plugging is 0 for e^2x we get: lim x->0 f(x)-f(0)/x-0 which is thr expression above

@thepotato1232 3 ай бұрын

@@t.b.4923 I see now,thank you

@EmmanuelGiouvanopoulos 2 ай бұрын

L'Hopital's rule is haunting this man.

@joshuahillerup4290 2 ай бұрын

Has there been a video showing where the limit of a product exists, but the product of the limits does not exist?

@bprpcalculusbasics 2 ай бұрын

You can just use 1/x and x as x approaches 0.

@aubertducharmont 3 ай бұрын

My way: set x=y, then use the l'Hopital's rule. It is probably not mathematically correct, but it works, if both x and y are aproaching the same value.

@projectseven2727 2 ай бұрын

Just a problem: there's a possibility that the limit does not exist. The issue with multivariable derivatives is that there are infinite directional limits and not just left and right sided. It is possible that limit x=y is not the same as lim x=0 or y=0 etc

@davidcroft95 3 ай бұрын

Is there an equivalent theorem of L'Hopital rule in calc 3?

@anthonyflanders1347 3 ай бұрын

Unfortunately no

@davidcroft95 3 ай бұрын

@@anthonyflanders1347 shucks! Thanks for the answer tho :)

@l9day 3 ай бұрын

Now number 5, the larch.... the larch

@איתיריכרדסון 3 ай бұрын

He wrote squezze lol

@chonkycat123 3 ай бұрын

Why can’t L’Hopital’s rule be used for the substitution method? It’s single-variable?

@griffinf8469 2 ай бұрын

You can use L’Hopital’s rule and you’ll get the same answer. The derivative of sin(t) is cos(t) and the derivative of t is 1. Plug in 0 to t and you’ll see that cos(0) is 1.

@chonkycat123 2 ай бұрын

@@griffinf8469 yeah, that’s how I would approach it. Just curious as to why bprp said you couldn’t use it

@chrisyoutube08 2 ай бұрын

@@chonkycat123he doesn't like using it for some reason, I've noticed it in other videos as well

@gileadedetogni9054 2 ай бұрын

Hi man, I can explain to you. The reason is that the limit of sin(h)/h as h goes to 0 appears on the derivative of sin(x) by definition. So, to know that the derivative of sin(x) is cos(x), you need to solve that limit first, resulting in a cyclic thing, get it?

@Brid727 2 ай бұрын

@@griffinf8469 yeah for sure, but what if you were to do the derivative of sin(t) using the limit definition? You would find this limit while working it out, and then you'll be tempted to use L'Hopital's rule. But that just doesn't make sense because you are taking tyhe derivative of sin(t) in the first place. That's why you have to find other ways to do it, preferably the squeeze/sandwich theorem