How To Evaluate Limits of Radical Functions | Calculus

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This Calculus video tutorial explains how to evaluate limits with radical functions such as square root functions.
Introduction to Limits: • Calculus 1 - Introduct...
How To Evaluate Limits From a Graph:
• How To Evaluate Limits...
Evaluating Limits By Factoring:
• Evaluating Limits By F...
Limits of Rational Functions :
• Limits of Rational Fun...
Limits of Radical Functions:
• How To Evaluate Limits...
Limits of Trigonometric Functions:
• Limits of Trigonometri...
___________________________________
How To Find The Limit at Infinity:
• How To Find The Limit ...
Infinite Limits + Vertical Asymptotes:
• Infinite Limits and Ve...
Limits at Infinity With Radicals:
• Limits at Infinity Wit...
Limits of Absolute Value Functions:
• Limits and Absolute Value
Limits of Composite Functions:
• How To Find The Limit ...
The Squeeze Theorem:
• Squeeze Theorem
____________________________________
Limits and Logarithms:
• Limits and Logarithms
Limits of Exponential Functions:
• Limits of Exponential ...
Piecewise Functions - Limits:
• Piecewise Functions - ...
3 Step Continuity Test:
• 3 Step Continuity Test...
Continuity and Differentiability:
• Continuity and Differe...
Limits - Test Review:
• One Sided Limits, Grap...
_____________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections
Limits - Free Formula Sheet:
bit.ly/3T3dD2X

Опубликовано:

2 окт 2024

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Комментарии : 84

@TheOrganicChemistryTutor Год назад

Limits - Free Formula Sheet: bit.ly/3T3dD2X Final Exams and Video Playlists: www.video-tutor.net/ Next Video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-wFeh4ByT0xs.html

@ashycallum Год назад

Question: How do you know when to take the conjugate of the numerator or the denominator?

@vananh3591 4 месяца назад

from my experience, in this type of equation , we always take the conjugate of denominator.

@thabangtibla 4 месяца назад

@@vananh3591 yes, because the denominator at this point is going to be zero, from the original expression

@prod_EYES 4 месяца назад

If you sub in the limit value and get zero on either one, u must take the conjugate of the one that got zero. In this case both were zero so he took the conjugate of both

@uvaldoandIsaaclopez6745 4 года назад

Dude you are by far the most excellent teacher and I’m blown away by the vastness of your knowledge! Thanks man!!!

@kevingarita386 Год назад

Great teacher! thanks a lot brother, I spent like 2 hours trying to find examples like these.

@georgesadler7830 Год назад

MR. Organic Chemistry Tutor, thank you for Evaluating Limits of Radical Functions using algebra tricks and tools. Limits problems in Calculus uses a heavy dose of algebra.

@gregoryg.philips2440 3 года назад

Thank you so much. I was having a little difficulty with my math 203 assignment but your explanation has given me a clearer understanding of the problem. May God bless you. 🙏

@ladymichellebarlisan7117 3 года назад

this example is quiet hard:( . In some parts ,the fraction was multiplied with the numerator and other parts were multiplied with the denominator

@lama907 3 года назад

frustrating isn't it, just when think you're getting the hang of it they change the approach leaving me dumbfounded

@yusiferzendric1489 3 года назад

@@lama907 oh it's not that hard. It's only little tricky calculation, you need to focus on that as a senior class student. And if you have problems in finding limits through calculations and factorization then you can try finding the derivative it is in fact much easier then this.

@JKMath 2 года назад

Determining whether to multiply by the numerator or the denominator for limits like these is all about where the square root is located in the function. In most cases if you see the radical/square root in the numerator you will want to multiply by the numerator. And if you see the radical/square root in the denominator you will want to multiply by the denominator. It is rare that you will see a radical in both the numerator and the denominator at the same time, but if you do, you can use try using either one.

@sissyime8004 2 года назад

even though you multiply it with the conjugate of the numerator first you will still get the same answer. it doesnt matter which one you use first

@YongHaoF 8 месяцев назад

@@sissyime8004 sorry but I multiply it with conjugate of numerator, I get the answer 0/0 (math error)

@deathmonk1409 4 года назад

I'm watching this because im bored

@joraii5913 4 года назад

SANA ALL

@malakayman2550 3 года назад

I watching this because I have a test

@dylanbinu10352 3 года назад

I’m not. I have homework due lol

@hdvvlakbd3407 Год назад

Wow

@fredartson 9 месяцев назад

Im watching this becuz im trying to finish calc before tenth

@KingGregforever Год назад

Thanks a lot 🙏 but is it necessarily compulsory to multiply top and bottom with the conjugate with the denominator?

@freeguylota Год назад

Are we meant to multiply by the numerators conjugate or the denominators own?

@Algebrainiac Год назад

This was a tough one! Great job!

@lucasemanuel8305 4 года назад

"oops"

@RaffaelloLorenzusSayde 2 года назад

Is multiplying by conjugate strictly for square roots?

@JKMath 2 года назад

In terms of methods for solving limits, yes. If you have a limit of a function with a square root, rationalizing by multiplying by the conjugate is the best way to go. There are better methods for solving limits of other kinds of functions without square roots, such as factoring if you have polynomials, or getting a common denominator if you have lots of fractions. Hope this helps!

@potchi3086 2 года назад

How about cube roots?

@efloof9314 Год назад

Calculus got me watching ts on my breaks

@akilanmc719 4 года назад

dude make a video about poisson's ratio

@Ivan-qv5xh 4 года назад

are you able to explain how to show the limit of 1/x as x approaches 0 does not exist using the e-d definition? I feel like it would be pretty good to look at since the function seems so simple yet so difficult to prove DNE, i've been stuck on it for so long.

@jimmykornelijegunnarsson4265 4 года назад

Use contradiction. If 1/x approach L as x goes to 0 (try pos and neg 0) then it should be expressed as abs(x) abs(1/x - L) < e.

@jerrychan4052 3 года назад

u r awesome, man

@unkownlast4907 Год назад

Why did (√7-x)(+2) equalled to +2√7-x ?

@jadeanndeloso6973 4 года назад

I learned a lot. Thanks!

@hluu01 4 года назад

Can you also have used l'hopital's rule?

@lawryvolcanoflames6293 4 года назад

hluu yep.

@snowlites 4 года назад

Living it up with Remote Learning

@alphasaffi9230 2 года назад

My mentor thank you very much may God 🙏 bless you sir.

@makishin3359 Год назад

thank you!!!

@Romeo-qk8tk 5 месяцев назад

Just use L'Hôpital! What is this, man? 😂 Although, multiplying by conjugates can resurrect indeterminate forms. I understand....

@rupom_1670 3 месяца назад

okay but it means that the answer is *not* 2/3 but rather something more accurate as it goes closer and closer to 2/3? because if we plug x=3 anywhere in the equation (doesn't matter at the first or at 6:07) we get 0/0 at the very first moment of the question which is indeterminable

@phathiscreations8721 3 года назад

So grateful for you. Thank you so much. That was of great help...🌸

@CaptainEchoPH 4 года назад

In the first example, can you instead use the numerator as a conjugate or should it always be the conjugate of the denominator when there are radicals in a fraction?

@usctrojanfreak 4 года назад

I tried it with the conjugate of the numerator and got the same answer. In fact, I got to the answer in fewer steps

@rolandoramos8498 3 года назад

SAME QUESTION. IT WAS ARDUOUS FOR ME FO FACTOR IT THOUGH. THUS, I JUST TOOK THE SAME STEPS AS HE DID.

@kimanhchung8372 4 года назад

Thanks

@malakayman2550 3 года назад

Your the best my teacher is way too fast

@salvadormedina4069 4 года назад

L’hopitals rule?

@johnlouiecaluminga3250 3 года назад

BSEE 1-4

@amarachimcjaphet6025 Год назад

please why did we divide by the denominator instead of the numerator in this problem

@andyguan3553 2 года назад

Thanks, helped alot!!

@jan-willemreens9010 Год назад

... An alternative way of solving this limit as follows: lim(x-->3)((sqrt(12 - x) - 3)/(sqrt(7 - x) - 2)) , Multiply the limit by ((7 - x) - 4)/((12 - x) - 9) (= 1/1) and split the limit in two limits according to the multiplication limit law: lim(AB) =lim(A)lim(B) as follows: [ lim(x-->3)((sqrt(12 - x) - 3)/((12 - x) - 9)) ][ lim(x-->3)((7 - x) - 4)/(sqrt(7 - x) - 2)) ] , Treat (12 - x) - 9 and (7 - x) - 4 as differences of two squares: (12 - x) - 9 = (sqrt(12 - x) - 3)(sqrt(12 - x) + 3) and (7 - x) - 4 = (sqrt(7 - x) - 2)(sqrt(7 - x) + 2) , After cancelling the common factors (sqrt(12 - x) - 3) and (sqrt(7 - x) - 2) of respectively limit #1 and #2 we obtain the final solvable limit form of: [ lim(x-->3)(1/(sqrt(12 - x) + 3)) ][ lim(x-->3)(sqrt(7 - x) + 2) ] = [ 1/(3 + 3) ][ 2 + 2 ] = (1/6)(4) = 4/6 = 2/3 ... Hoping this way is also appreciated a bit (less algebra)... Thank you for all your valuable math efforts, Jan-W

@Cloud7050 2 года назад

What sorcery is this? Lol I was confused by a similar question involving the replacement rule, limits, and square roots. Thanks.

@_wash 2 года назад

You are literally a legend omg

@shalian5687 3 года назад

Seems like it switched from numerator to denominator then denominator to numerator and change the sign instead of -3 is +3 then -2 to +2 then direct substitution. Idk I just noticed. correct me if I'm wrong. Hihi