Multiplying by the conjugate [Example] [Algebraic limits]

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Multiplying by the conjugate [Example] [Algebraic limits]

Рет қаралды 12,424

11 ай бұрын

Worked example for solving a limit by the technique of multiplying by the conjugate.
Original content © 2023 Jung-Lynn Jonathan Yang
CC-BY-NC-ND
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Пікірлер: 40

@xadxtya 6 ай бұрын

Wow I really liked that little flashcard for Conjugate

@eboroemmanuel5606 2 ай бұрын

Wow i love this video please i need more even on physics and chemistry

@chethkeathmaths 5 ай бұрын

The best video

@MegumiKato418 6 ай бұрын

Thank you

@surendrakverma555 4 ай бұрын

Good

@pkfrozen7867 6 ай бұрын

great video!

@bouazabachir4286 6 ай бұрын

Thanks a lot professor I follow you from Algeria.

@DrYheartLab 6 ай бұрын

I am not officially a professor because I am unemployed. I do things that professors do, like teaching, supervising undergraduate/graduate research projects, and evaluating funding applications. Thank you for watching!

@jan-willemreens9010 7 ай бұрын

... Good day, Observing the expression of the limit ( SQRT(X + 2) - 3 ) / ( X - 7 ), I see the occasion to rewrite the denominator ( X - 7 ) as follows ... X - 7 = ( X + 2 ) - 9 , and treat ( X + 2 ) - 9 as a " Difference of 2 Squares " ... ( X + 2 ) - 9 = [ SQRT(X + 2) - 3 ] * [ SQRT(X + 2) + 3 ] ... now recognizing a common factor between numerator and denominator expression of [ SQRT(X + 2) - 3 ] ... which simplifies the original limit as .... LIM(X->7)[ 1 / ( SQRT(X + 2) + 3 ) ] = 1 / ( SQRT(9) + 3 ) = 1 / ( 3 + 3 ) = 1/6 ... solving this limit exercise by applying " Factoring the Denominator " ... thanking you DrY(black heart)Lab for your instructive presentation .... best regards, Jan-W

@DrYheartLab 7 ай бұрын

Amazing, @Jan-W, I could not have thought of your solution! Absolutely, write the “x − 7” as “(x + 2) − 9” and then factor as a difference of squares to cancel with the numerator. I am glad you brought up this alternative solution! 💡

@ximer3316 6 ай бұрын

Thats splendid. I would not be able to come up with that in 1000 years!!

@DrYheartLab 6 ай бұрын

Not to worry! There may be multiple ways to solve limits. Just use what works well and accurately for you.

@victorpaesplinio2865 3 ай бұрын

"Why don't use L'Hopital?", one may ask. Well, it depends on the purpose. Since you need to learn limits in order to learn derivatives in a standard calculus 1 course, you will face many 0/0 limits where you will be required to solve algebraically before being introduced to the concept of derivatives. Every year at my college I face myself trying to help new students to manipulate a limit where L'Hopital would solve in two lines, just because they are in the beginning of the course haha

@DrYheartLab 3 ай бұрын

This is exactly what I have to do for teaching calculus! It's neat that we use limits to evaluate derivatives and then use derivatives, in the form of l'Hôpital's Rule, to solve limits.

@alban6423 6 ай бұрын

It is the literal definition of the derivative of f(x) = sqrt(x+2) at x = 7

@DrYheartLab 6 ай бұрын

Good call! It is! I think this is why calculus courses begin with limits because then we can extend limits to the limit definition of the derivative.

@NathanielSundayHaruna 3 ай бұрын

Good evening ask questions

@philippedelaveau528 6 ай бұрын

But since the limit for 7 of numerator and denominator are both equal to zero, L”Hospital’s rule works very well.

@DrYheartLab 6 ай бұрын

You are right! L'Hôpital's Rule is perfect for indeterminate forms “0/0” (the situation here) and “±∞/±∞”. In the calculus courses I took and taught, we had to solve limits algebraically. On my midterm exam, the instruction specifically told us to not use L'Hôpital's Rule. It was not until later on in the course, about applications of derivatives, that we had the option of L'Hôpital's Rule.

@philippedelaveau528 6 ай бұрын

Thank you. Sorry for the advance@@DrYheartLab

@DrYheartLab 6 ай бұрын

I am glad you brought up l’Hôpital’s Rule! If the question does not specify which technique to use, or not to use, l’Hôpital’s Rule is often a faster method than algebra.

@NathanielSundayHaruna 3 ай бұрын

If Y=1/2√x

@DrYheartLab 3 ай бұрын

Sorry, I did not understand your question! Could you tell me more?

@shivx3295 6 ай бұрын

Did it orally by just rationalising

@DrYheartLab 6 ай бұрын

That is great to be able to solve this question mentally (by inspection)! Solving mentally is especially suitable on questions where it is not necessary to show calculation work, like multiple choice questions.

@omnamonarayan3523 4 ай бұрын

🎉hey 👋🏻☺️👋🏻😊😌🤗@

@user-vf2di7hx7h 4 ай бұрын

6^-1

@DrYheartLab 4 ай бұрын

That is the correct answer!

@zzzz-yw4zk 6 ай бұрын

Couldnt u just do l'hopital?

@DrYheartLab 6 ай бұрын

Absolutely, l’Hôpital’s Rule is another way to solve this limit question! In fact, l’Hôpital’s Rule is applicable for limits of the form “0/0” (what we have here) and “±∞/±∞”. In the beginning lessons and midterm(s) in Calculus I, instructors tend to tell students to solve limits using algebraic methods rather than l’Hôpital’s Rule. However, if the question does not specify which method to use, l’Hôpital’s Rule can be an easier and faster method than algebra.

@user-yh9iq9tn5k 4 ай бұрын

meanwhile i'm just using l'hopital rule

@DrYheartLab 4 ай бұрын

Yes, use l’Hôpital's Rule if the question does not tell you not to use it! In the calculus courses I took and have to teach, we could only use algebraic techniques and were not allowed to use l’Hôpital's Rule in the unit about limits.

@Will-nf9gf 12 күн бұрын

I think actually the conjugate is way faster because of the x - 7 cancellation but I like how l'hopital is more general so honestly me too

@DrYheartLab 10 күн бұрын

Neat! Multiplying by the conjugate may be faster because the root differentiates by the Chain Rule, which brings in more factors to deal with and makes l’Hôpital's Rule a bit complicated. Alternatively, we can try a mix of both methods! Start with multiplying by the conjugate, then as long as the indeterminate form is appropriate (±∞/±∞ or 0/0), use l’Hôpital's Rule. Whatever method you prefer, use the method that works fastest and most accurately for you and is also allowed by the exam instructions.

@user-er6zr1tm3i 3 ай бұрын

? Bernulli Lopital is faster!

@DrYheartLab 3 ай бұрын

In the first few lessons of Calculus I, we are told to solve limits algebraically and do not use l’Hôpital's Rule. Although, if the teacher does not specify which method to use, or not to use, then yes, use l’Hôpital's Rule because it is faster!

@elmeralvarado1236 6 ай бұрын

Why didn't you conjugate (x-7) with (x+7)?

@DrYheartLab 6 ай бұрын

Good question! When we multiply (x − 7) and (x + 7), we have the difference of squares x² − 49, which does not cancel much. Instead, if we target the awkward square root, then we can eliminate the square root as a difference of squares.

@pauselab5569 6 ай бұрын

that is way to specific. no way anyone would prefer this over l'hopital which is both easier to use and more general.

@DrYheartLab 4 ай бұрын

There is a lot of annoying algebra! For some reason, I had to learn this as a student and is also in the curriculum I teach in multiple universities. L'Hôpital's Rule is easier and faster.