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. 🙏
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
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.
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.
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?
@@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.
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.
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!
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
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
... 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
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Dude you are by far the most excellent teacher and I’m blown away by the vastness of your knowledge! Thanks man!!!
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. 🙏
Question: How do you know when to take the conjugate of the numerator or the denominator?
from my experience, in this type of equation , we always take the conjugate of denominator.
@@vananh3591 yes, because the denominator at this point is going to be zero, from the original expression
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
Great teacher! thanks a lot brother, I spent like 2 hours trying to find examples like these.
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.
This was a tough one! Great job!
Thanks a lot 🙏 but is it necessarily compulsory to multiply top and bottom with the conjugate with the denominator?
My mentor thank you very much may God 🙏 bless you sir.
You are literally a legend omg
So grateful for you. Thank you so much. That was of great help...🌸
THANK YOU SO MUCH MAN 😭
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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.
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.
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?
I tried it with the conjugate of the numerator and got the same answer. In fact, I got to the answer in fewer steps
SAME QUESTION. IT WAS ARDUOUS FOR ME FO FACTOR IT THOUGH. THUS, I JUST TOOK THE SAME STEPS AS HE DID.
Are we meant to multiply by the numerators conjugate or the denominators own?
Thanks, helped alot!!
What sorcery is this? Lol I was confused by a similar question involving the replacement rule, limits, and square roots. Thanks.
this example is quiet hard:( . In some parts ,the fraction was multiplied with the numerator and other parts were multiplied with the denominator
frustrating isn't it, just when think you're getting the hang of it they change the approach leaving me dumbfounded
@@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.
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.
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
@@sissyime8004 sorry but I multiply it with conjugate of numerator, I get the answer 0/0 (math error)
u r awesome, man
Thanks a lot, sir.
Is multiplying by conjugate strictly for square roots?
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!
How about cube roots?
Can you also have used l'hopital's rule?
hluu yep.
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
Thanks
omg thank you so much!!
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
Why did (√7-x)(+2) equalled to +2√7-x ?
I learned a lot. Thanks!
thank you!!!
Thanks sir
... 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
Thanks Sir 👍
please why did we divide by the denominator instead of the numerator in this problem
Calculus got me watching ts on my breaks
dude make a video about poisson's ratio
Your the best my teacher is way too fast
I'm watching this because im bored
SANA ALL
I watching this because I have a test
I’m not. I have homework due lol
Wow
Im watching this becuz im trying to finish calc before tenth
BSEE 1-4
L’hopitals rule?
(lim)┬(x→16)〖(x^2-256)/( 4-√x)〗
-256 . I'm late hahaha
"oops"
i had a very similar problem and i was trying to factor so bad
so lesson learned follow the path of less resistance
Why don't use L'Hopital's method??
most people learn limits before derivatives
5:03
G 183
9
I have a faster method on that sir. I can evaluate as faster as 3 seconds of any limits conjugate.
how is it brethren? would you mind if you share it?
Just use L'Hôpital! What is this, man? 😂
Although, multiplying by conjugates can resurrect indeterminate forms. I understand....
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Are you god
Second
6:41 so extra
"How can we..".. you pronounce it so differently..
Fk its so hard huhu
Medium it is not perfect like other video
First view first. Comment
messy, i can't understand it well, sorry
then you should have skipped it
Thanks sir