Evaluating Limits Algebraically: Radicals, Factoring, Fractions & More! || Calculus 1

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Hey everyone! This video will teach you the basics of evaluating limits algebraically. We'll talk about dealing with radicals, having to factor, finding common denominators in limits, pretty much everything and anything to get you started with this!
Timestamps:
01:53 - When Limits Get Interesting
07:00 - Indeterminate Cases
09:25 - Factoring
17:01 - Simplifying (Common Denominators, Expanding Polynomials)
29:47 - Rationalizing (Dealing w/ Radicals)
Intro to Limits PDF: COMING SOON (sorry, just need to setup my website first)
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Опубликовано:

28 сен 2024

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Комментарии : 10

@LudusYT 4 года назад

Timestamps & Donations: 01:53 - When Limits Get Interesting 07:00 - Indeterminate Cases 09:25 - Factoring 17:01 - Simplifying (Common Denominators, Expanding Polynomials) 29:47 - Rationalizing (Dealing w/ Radicals) If this video helped, consider supporting me on Patreon!!! My Patreon: www.patreon.com/ludus1

@zaza8870 4 года назад

I’m not American so I don’t know if you use this but l'hopital's rule is I think an easier method when finding limits.

@LudusYT 4 года назад

Yes, we learn l’hosptial’s rule, but you generally learn limits before derivatives, and so before you learn l’hospital’s rule, you learn this.

@zaza8870 4 года назад

Ludus, that makes sense. I was taught the other way. Thanks for clearing it up for me

@LudusYT 4 года назад

No prob! L’hospital’s rule is definitely more convenient, but since you generally learn limits before derivatives, students can’t use that rule yet. I know that’s not always the order in which they are taught though!

@seamus7755 4 года назад

you're the best! Just curious what college did you go to or are attending right now?

@layzaramirez5192 4 года назад

I’m still confused on how to factor with the abc method😐 I have 3x squared minus 2x minus 8.

@jan-willemreens9010 Год назад

...An alternative way to solve your limit at about 30:00: lim(x-->0)((sqrt(x^2 + 9) - 3)/x^2) (rewrite the denominator x^2 to match the numerator as follows: x^2 = (x^2 + 9) - 9 and treat this new expression as a difference of two squares as follows: x^2 = (x^2 + 9) - 9 = (sqrt(x^2 + 9) - 3)(sqrt(x^2 + 9) + 3), and finally cancelling the common factor (sqrt(x^2 + 9) - 3) of the numerator and denominator to obtain the following solvable limit form): lim(x-->0)(1/(sqrt(x^2 + 9) + 3) = 1/(3 + 3) = 1/6... faster and less chance of making errors; people often focus to quickly on the root part while in the non root part of the problem a lot of valuable information is available... Thank you Ludus for your honest math efforts, Jan-W