Wow I really liked that little flashcard for Conjugate
@eboroemmanuel56062 ай бұрын
Wow i love this video please i need more even on physics and chemistry
@chethkeathmaths5 ай бұрын
The best video
@MegumiKato4186 ай бұрын
Thank you
@surendrakverma5554 ай бұрын
Good
@pkfrozen78676 ай бұрын
great video!
@bouazabachir42866 ай бұрын
Thanks a lot professor I follow you from Algeria.
@DrYheartLab6 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-willemreens90107 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
@DrYheartLab7 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! 💡
@ximer33166 ай бұрын
Thats splendid. I would not be able to come up with that in 1000 years!!
@DrYheartLab6 ай бұрын
Not to worry! There may be multiple ways to solve limits. Just use what works well and accurately for you.
@victorpaesplinio28653 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
@DrYheartLab3 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.
@alban64236 ай бұрын
It is the literal definition of the derivative of f(x) = sqrt(x+2) at x = 7
@DrYheartLab6 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.
@NathanielSundayHaruna3 ай бұрын
Good evening ask questions
@philippedelaveau5286 ай бұрын
But since the limit for 7 of numerator and denominator are both equal to zero, L”Hospital’s rule works very well.
@DrYheartLab6 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.
@philippedelaveau5286 ай бұрын
Thank you. Sorry for the advance@@DrYheartLab
@DrYheartLab6 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.
@NathanielSundayHaruna3 ай бұрын
If Y=1/2√x
@DrYheartLab3 ай бұрын
Sorry, I did not understand your question! Could you tell me more?
@shivx32956 ай бұрын
Did it orally by just rationalising
@DrYheartLab6 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.
@omnamonarayan35234 ай бұрын
🎉hey 👋🏻☺️👋🏻😊😌🤗@
@user-vf2di7hx7h4 ай бұрын
6^-1
@DrYheartLab4 ай бұрын
That is the correct answer!
@zzzz-yw4zk6 ай бұрын
Couldnt u just do l'hopital?
@DrYheartLab6 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-yh9iq9tn5k4 ай бұрын
meanwhile i'm just using l'hopital rule
@DrYheartLab4 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-nf9gf12 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
@DrYheartLab10 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-er6zr1tm3i3 ай бұрын
? Bernulli Lopital is faster!
@DrYheartLab3 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!
@elmeralvarado12366 ай бұрын
Why didn't you conjugate (x-7) with (x+7)?
@DrYheartLab6 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.
@pauselab55696 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.
@DrYheartLab4 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.