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

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...
Get a derivative t-shirt: 👉 bit.ly/derivat...
Use "WELCOME10" for 10% off
Subscribe for more precalculus & calculus tutorials 👉 ‪@bprpcalculusbasics‬
-------------------
If you find this channel helpful and want to support it, then you can
join the channel membership and have your name in the video descriptions:
👉bit.ly/joinjus...
buy a math shirt or a hoodie (10% off with the code "WELCOME10"):
👉 bit.ly/bprp_merch
I use these markers 👉 amzn.to/3skwj1E
-------------------
😊 Thanks to all channel members 😊
Sandglass Dªrksun Seth Morris Andrea Mele
---------------------------------------------------------
"Just Calculus" is dedicated to helping students who are taking precalculus, AP calculus, GCSE, A-Level, year 12 maths, college calculus, or high school calculus. Topics include functions, limits, indeterminate forms, derivatives, and their applications, integration techniques and their applications, separable differential equations, sequences, series convergence test, power series a lot more. Feel free to leave calculus questions in the comment section and subscribe for future videos 👉 bit.ly/just_calc
---------------------------------------------------------
Best wishes to you,
#justcalculus

[@catlooks]

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

[@monasimp87]

Proof by intimidation lol

[@santoriomaker69]

proof by shirt design

[@leviuchiha3706]

How to get this shirt

[@youz123]

source: It's on my shirt

[@smugless191]

0^0 is also on his shirt.

[@nerduto1]

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

[@theRipintheChat]

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

[@vaxjoaberg]

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]

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]

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]

@@Goldfrapplol Je suis corrigé.

[@vaxjoaberg]

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

[@hassanalihusseini1717]

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

[@azzteke]

L´Hopital

[@robinson5923]

Is that the Heisenberg uncertainty cat in your shirt?

[@EE-ho1iz]

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

[@oenrn]

You mean Schrödinger?

[@GirishManjunathMusic]

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]

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

[@Rzko]

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]

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

[@GirishManjunathMusic]

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

[@GirishManjunathMusic]

@@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.

[@black_pantheon]

watching your videos actually inspires me to buy a whiteboard with markers and a eraser to study math lol no joke

[@channelsixtysix066]

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

[@pan_nekdo]

I prefer u=1/x substitution.

[@leviuchiha3706]

Doesn't make any difference.. as it becomes lim(u-infinity) (1/u)^(1/u)^1/2
... it becomes more complicated....

[@pan_nekdo]

@@leviuchiha3706 After taking logaritm it's the L'H use obvious.

[@DaltonPritt]

Thank you!

[@mutenfuyael3461]

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))

[@kepler4192]

yes exactly

[@smitad7881]

Thanks. Prefer 1st method.

[@chickenadobo9966]

natatawa talaga ako pag nagssigh siya HAHAHHAHA (filipino here struggling sa engineering)

[@domc3743]

Let x= y^2 then we have y^2y = lim y^y * lim y^y = 1*1 =1

[@absurdtuber3341]

Thank you

[@space_engineer17]

0:10
Why 0^0 is indeterminate?
Because it's on the shirt!!😂
Notice that Schrodinger cat
Indeterminate cat😱

[@NXT_LVL_DVL]

You did the same method two times actually

[@saveerjain6833]

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

[@ZipplyZane]

Ln is a *continuous* function.

[@saveerjain6833]

@@ZipplyZane thank you!

[@ZipplyZane]

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

[@saveerjain6833]

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

[@matteocilla9482]

hello, i forget the conditions to use L’H ?? can you remind me pls ??

[@tricky778]

1 requirement is that the numerator and denominator either both approach zero or both approach infinite size

[@tricky778]

Another is that both are differentiable in the relevant region

[@Measure_differentiable]

This is wrong. YOU CANT USE L'HOPITALS RULE HERE. MOREOVER YOU DO NOT NEED TO .

[@AKeenCabinTV]

This was just recommended to me but wtf is this...