please use L'Hospital's Rule for the limit of x^sqrt(x) as x goes to 0+

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We will use L'Hospital's Rule for the limit of x^sqrt(x) as x goes to 0+. Even though we will get a 0^0 when we plug in 0 into x^sqrt(x), 0^0 IS an indeterminate form so we must do more work in order to determine the limit. Here's an example of the limit with the indeterminate form 0^0 but we do not get 1. 👉 • a 0^0 limit that appro...
This is how to spell L'Hospital's rule: 👉 • how to spell L’Hosptis...
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9 апр 2022

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Комментарии : 52

@catlooks 2 года назад

"This is an indeterminable form because it's on my shirt" nice proof lol

@monasimp87 2 года назад

Proof by intimidation lol

@santoriomaker69 2 года назад

proof by shirt design

@leviuchiha3706 2 года назад

How to get this shirt

@youz123 Год назад

source: It's on my shirt

@smugless191 5 месяцев назад

0^0 is also on his shirt.

@nerduto1 2 года назад

How to always be correct in math: Use your shirt as proof.

@theRipintheChat Год назад

I find it hilarious that you have stocked up a shelf full of expo markers like your holding out for the apocalypse.

@vaxjoaberg 2 года назад

Poor L'Hopital. Because we think it's funny an entire generation of math enthusiasts are going to grow up believing his name is L'Hospital. And in a generation after that his original name will be lost to time.

@ericsills6484 2 года назад

He actually made a mistake. There's no 's' in L'Hopital. I guess I can give him a break though. He wasn't around in the 1600's :-)

@Goldfrapplol 2 года назад

Well, actually in older French spelling his name was spelled L'Hospital, which is why it's written today in modern French with an o circumflex: L'Hôpital. Source: Math major currently speaking French at A1 level so I can understand fundamental math treatises. 😀

@ericsills6484 2 года назад

@@Goldfrapplol Je suis corrigé.

@vaxjoaberg 2 года назад

@@Goldfrapplol Thanks, that's good to know. I feel less bad for poor L'Ho(s)pital, now.

@robinson5923 2 года назад

Is that the Heisenberg uncertainty cat in your shirt?

@EE-ho1iz 2 года назад

It's all the indeterminate form, so it's technically an indeterminate cat! Wait a dang minute... :O

@oenrn 2 года назад

You mean Schrödinger?

@hassanalihusseini1717 2 года назад

Nice example for the use of Hopital rule! Thank you!

@azzteke Год назад

L´Hopital

@DaltonPritt 3 месяца назад

Thank you!

@absurdtuber3341 Год назад

Thank you

@smitad7881 Год назад

Thanks. Prefer 1st method.

@saveerjain6833 2 года назад

1:12 Sorry what word did you use for the type of function that natural log is?

@ZipplyZane 2 года назад

Ln is a *continuous* function.

@saveerjain6833 2 года назад

@@ZipplyZane thank you!

@ZipplyZane 2 года назад

@@saveerjain6833 To be clear, it only works because ln is continuous in the relevant interval, which is near 0 on the positive side.

@saveerjain6833 2 года назад

@@ZipplyZane no yeha i get that just didn’t hear the word

@channelsixtysix066 2 года назад

One other important thing to remember, all you mathematicians out there. If it's on the t-shirt, it gotta be right.

@GirishManjunathMusic 2 года назад

Before watching the video: Find the lim(x→0+) x↑(x↑½) Consider x↑(x↑½) as exp((x↑½)(lnx)) lim(x→0+) exp((x↑½)(lnx)) as exponential function is continuous over the target domain, limit of the exponential function of a function of x is the same as the exponential function of the limit of that function of x. thus lim(x→0+) exp((x↑½)(lnx)) = exp(lim(x→0+) ((x↑½)(lnx))) Defining L = lim(x→0+) ((x↑½)(lnx)): The question now reduces to: Find exp(L). L = lim(x→0+) ((x↑½)(lnx)) This is of the indeterminate form 0·(-∞) Rewriting to obtain a more workable indeterminate form, L = lim(x→0+) ((lnx)/(x↑(-½))), which is of the indeterminate form (-∞)/∞, allowing for the use of L'Hospital's Rule. L = lim(x→0+) (d/dx (lnx)/d/dx (x↑(-½))), = lim(x→0+) ((x↑(-1)/((-½)(x↑(-3/2)))) = lim(x→0+) (-2√x) = 0 L = 0. exp(L) = exp(0) = 1. Thus, lim(x→0+) x↑(x↑½) = exp(L) = 1.

@GirishManjunathMusic 2 года назад

After Watching the Video: hey my solution was essentially the same as the second method you showed!

@Rzko 2 года назад

Why do americans love this "L'hopitals rule"? Just use the "growth comparison" (that's how it's named in french), α^x is dominant over x^α which is dominant over ln(x), with "α" a constant

@tricky778 2 года назад

Fyi, the up arrow normally denotes the Boolean 'nand' connective

@GirishManjunathMusic 2 года назад

@@Rzko this is a series on L'Hospital's Rule in finding limits of certain indeterminate forms.

@GirishManjunathMusic 2 года назад

@@tricky778 FYI, as we're working with limits and not Boolean logic, the up arrow denotes exponentiation. a↑b = a multiplied by itself b times.

@mutenfuyael3461 2 года назад

If you know that xln(x) when x approches 0 is 0, you can write sqrt (x) * ln(x) as 2(sqrt(x)*ln(sqrt (x))= 2*0=0=L because sqrt (x) approches 0 when x approches 0, so do it sqrt (x)*ln(sqrt(x))