"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
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.
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!
... 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
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! 💡
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.
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.
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.
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.
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.
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!
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.
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.
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.
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"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
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.
Good evening ask questions
🎉hey 👋🏻☺️👋🏻😊😌🤗@
? Bernulli Lopital is faster!
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!
Wow I really liked that little flashcard for Conjugate
... 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
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! 💡
Thats splendid. I would not be able to come up with that in 1000 years!!
Not to worry! There may be multiple ways to solve limits. Just use what works well and accurately for you.
But since the limit for 7 of numerator and denominator are both equal to zero, L”Hospital’s rule works very well.
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.
Thank you. Sorry for the advance@@DrYheartLab
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.
meanwhile i'm just using l'hopital rule
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.
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
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.
If Y=1/2√x
Sorry, I did not understand your question! Could you tell me more?
Wow i love this video please i need more even on physics and chemistry
that is way to specific. no way anyone would prefer this over l'hopital which is both easier to use and more general.
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.
Good
The best video
It is the literal definition of the derivative of f(x) = sqrt(x+2) at x = 7
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.
Thanks a lot professor I follow you from Algeria.
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!
Couldnt u just do l'hopital?
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.
great video!
Did it orally by just rationalising
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.
Thank you
6^-1
That is the correct answer!
Why didn't you conjugate (x-7) with (x+7)?
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.