Math for fun#1, design your limit

Calculus 1 limit.

Math For Fun#1, Limit
math for fun series#1, limits, precalc, calculus, algebra.

[whiz]

That laugh right at the end there caught me off guard!

[JSSTyger]

b = 9, a = 2...b determined from 0/0 form.  "a" determined from L'Hopital's rule that the derivative of the numerator divided by the derivative of the denominator will equal the limit in the case of a 0/0 indeterminate form.  Good problem.

[weerman]

I love these Math For Fun series! Keep them going!

[tsujimasen]

From 2:52, Why not just use L'Hospital?

Would save the conjugate multiplication.

[markmajkowski]

1-Multiply both sides by 3x
2-add 3 to both sides
3-square both sides
4-limx0 9ax+9b=x^2+18x+81
5-equate terms as x>0
9a=18, 9b=81 2=a 9=b

[observ_]

I actually had exercises similar to this one in the finals, those are really fun for those who would like to challenge themselves.
Nice video, as always (although it is 3 years old it's still fun to watch)

[ntruesdale]

Cross multiply first. You aren't forced to take the limit right away. Multiplying by the conjugate is the right way to solve for a, but, once again, can be done after multiplying both sides by x first. You are right about 0/0 still giving a chance to the expression equaling 1/3, but that is not the only route to take.

[riptidemonzarc]

I did a heuristic calculation with algebra that resulted in an equation a9x + b9 = x^2 + 18x + 81. Disregarding x^2 as x -> 0 gives the appropriate values.

[KnakuanaRka]

Pretty simple. First, this has to be an indeterminate form, so the numerator must go to 0 along with the denominator; s(ax+b)-3=0 at x=0 means s(b)-3=0 and b=9.
Second, using L’H, the derivative of the top is a/2s(ax+9), which at 0 is a/2s(9), or a/6. Since the derivative of the bottom is equal to 1, this must be 1/3, to make the overall fraction 1/3. a/6=1/3 easily gets you a=2. Simpler than what you did with the conjugates.

[alextownsend]

L'Hospital's rule is great, but you still have to get rid of the x in the denominator and doing that requires simplification via conjugate multiplication which leads to difference of squares. Conjugate multiplication is necessary for verifying your answer. Plugging a, b, and x in at the end otherwise would not work because it would yield the indeterminate form of 0/0. If the x in the denominator is cancelled out then the limit does indeed approach 1/3. And as far as the quotient rule of limits is concerned the same issue arises being that solving works, but not checking your answer.

[gabrielfoos]

I did à Taylor expansion and got sqrt(b) had to be equal to 3 or else the form becomes indertimate (so b=9), and by looking at the last term you then have a/2sqrt(b)=1/3 by using b’s value and by isolating a you finally get a =2

[marouaniAymen]

Thanks for the video; actually I used another method. I considered the function f(x)=sqrt(ax+b)-3
in this case, the limit is definition of the derivative of f on 0. So to determine b, we use f(0)=0. Knowing b, we derermine a using f'(0)=1/3.

[riccardofroz]

A more general solution would be a^2=4b/9

[moazselim]

You could spend 5 minutes messing around with conjugation OR, you could just do it mentally in 30 seconds using L'Hopital with the assumption that the numerator is equal to 0. When I took a derivative, the denominator was 1 and the numerator was a/(2root(ax+b)) I set that equal to 1/3 and since I need a one in the numerator I just made a=2 so it would cancel with the bottom. Then since the limit approaches 0 I plugged in 0 and I got 1/root(b) = 1/3 and you can instantly tell b would be 9.

[tahsintarif]

Do this as "Math for Fun" : Prove the volume of Cone, Sphere without using calculus or any advanced theorem. (Algebra and Basic geometry only) .

[Cor]

I failed to understand your result by applying the limit definition. For some number e>0 there exist a number d>0 such that: |f(d)-1/3|<e for all 0<|x|<d. Suppose e=0.00001, then d=21.72... However, much bigger then 0.00001. Could someone exhibit a number e>0 and a number d such that |f(d)-1/3|<e for all 0<|x|<d?

[frede]

At 5:42, couldn't you also say that sqrt. 9 is -3?? If this is the case, you would get -3+3=0. Therefore, you would get a/0=1/3. In order for this not to be infinity, we have to set a=0. So a has two solutions, 0 and 2. This might be wrong, please explain why if that's the case. 😊

[waikeanng]

woww!I really like your video.😍keep it up!!!

[kamineko]

For the first part, why didnt you move the x onto the other side since it is possible because there is a limit which means that x isnt 0

[منهاجالعلموالثقافة]

Thanks for you from my heart I from Sadie Arabia 😝😘❤️