two nice math problems

[@goodplacetostop2973]

0:08 First problem 4:37 Second problem 11:29 Good Place To Stop

[@Ahmed-Youcef1959]

😀😀😀

[@mtaur4113]

at 8:50, "always negative" for *integer* x, due to the interval of positivity being (-1,0).

[@kavimahajan8412]

Yea

[@srikanthtupurani6316]

There are some proofs which don't use lhospital rule. But what they do is equivalent to using all the math needed in proving lhospital rule. It is an interesting problem.

[@wiseSYW]

it asks for a solution in the integers intead of natural numbers, and it has a cube function, so a high chance that the solutions are negative numbers

[@TheEternalVortex42]

8:40 The inequality is only true for integers. I guess that's implied since we are looking for solutions over integers, but the other inequality we derived worked for all real numbers.

[@jay_13875]

The solution [x=-1, y=0] is pretty easy to find by writing x³+x²+x+1 as (x⁴-1)/(x-1) and testing y=0. I didn't figure out a way to capitalize further on that observation though.

[@Happy_Abe]

Interestingly the first problem would imply that the derivative limits to 0. Which intuitively makes sense if f converges itself

[@MizardXYT]

A similar problem is y³ = x² + x + 1. This has a nontrivial solution of { x = 18, y = 7 }.

[@miraj2264]

I don't see the need to use (x-1)^3 as the lower bound. If you use x^3, you get x^3 0

[@liyi-hua2111]

4:34 lim f(x) = lim （f(x)+f’(x)/a). the function seems pretty restricted. Oh, because both limits exist, we can derive that lim f’(x)=0.

[@minamagdy4126]

For the first problem, if we assume that the limit of f(x) exists, then the condition given by the video's proof can only be false for that limit being zero. Also, in that case, the derivative limit just simplifies to the limit of f'(x) equalling L, which we can (I believe) prove also equals zero, thus validating that case as well. Only thing remaining is to prove that the limit of f(x) has to exist if the derivative limit does.

[@robshaw2639]

I just asked chatgpt to recommend a calculus textbook and was very pleased with the results considered “rigorous/advanced”: Apostol vol 1 and 2, Spivak, and (wow) Hardy! I’m in the Apostol camp, and I know Michael is in the Spivak camp, so this was very interesting, not to mention the 3rd choice was Hardy himself! Anyway,now I am curious to find this Spivak text for myself….

[@Merssedes]

For the first problem assumption about the limit should be just considered as special case.

[@alexchan4226]

the solution is constant and in linear function.

[@MacHooolahan]

Was slightly waiting for the wonderful moment these two results were brought together... :D

[@demenion3521]

too complicated solution for the second problem. don't use (x-1)³ for the lower bound, but just x³

[@Charliethephysicist]

The additional condition is not needed for the first problem. For arbitrary \epsilon>0, \exists x_0>0 i |e^{-ax}d/dx(e^{ax}f(x))-L|< \epsilon, \forall x>x_0. Then through algebra and integration, you can deduce the limit.

[@petver29]

There is no need to make an additional assumption. This statement follows from the chain rule and the provided limit.

[@bmschech]

I just tried the second problem on the new deepseek and here is what it did. pretty cool, no? To find all integer solutions to the equation \( y^3 = 1 + x + x^2 + x^3 \), we start by testing small integer values for \( x \): - **\( x = 0 \):** \( y^3 = 1 + 0 + 0 + 0 = 1 \), so \( y = 1 \). Solution: \( (0, 1) \). - **\( x = -1 \):** \( y^3 = 1 - 1 + 1 - 1 = 0 \), so \( y = 0 \). Solution: \( (-1, 0) \). For other values of \( x \), the right-hand side does not yield a perfect cube. To generalize, we factor the equation as \( y^3 = (x + 1)(x^2 + 1) \). Analyzing the greatest common divisor (GCD) of \( x + 1 \) and \( x^2 + 1 \): - If \( x \) is **even**, \( \gcd(x + 1, x^2 + 1) = 1 \). Both factors must be cubes: \( x + 1 = a^3 \) and \( x^2 + 1 = b^3 \). Testing small \( a \), only \( a = 1 \) works, giving \( x = 0 \). - If \( x \) is **odd**, \( \gcd(x + 1, x^2 + 1) = 2 \). Writing \( x + 1 = 2a^3 \) and \( x^2 + 1 = 2b^3 \), only \( a = 0 \) works, giving \( x = -1 \). Substituting \( y = x + k \) into the equation and solving the resulting quadratic shows that only \( k = 1 \) yields valid integer solutions, confirming \( x = 0 \) and \( x = -1 \). **Final Answer:** The integer solutions are \(\boxed{(-1, 0)}\) and \(\boxed{(0, 1)}\).

[@neuralwarp]

When are you going to start answering the question?

[@jay_13875]

I'm not sure what the point of the limit property (first problem) is. If the limit of f'(x) is 0, the relation is trivially true and won't help you find the limit of f(x). If the limit of f'(x) is +-infinity or any non-zero constant, the limit of f(x) has to be infinite. It could only be conceivably applicable if f'(x) is bounded but does not converge to a limit (i.e. oscillates around 0). But the example that I tested, f(x) = exp(-x)*sin(exp(x)), didn't even work since f'(x) = cos(exp(x))-f(x). The best (it seems) you could do is set a = 1, but then you're stuck with the limit of cos(exp(x)), which doesn't exist.

[@Rmsesii]

Hey, is there a way to solve the second problem with linear algebra? If I have the roots of a polynomial, those could be the kernel of my linear transformation, and then we can use the kernel to find the image? And somewhere in there lies the solution lol. Nice Use of the squeeze theorem tho...I think.

[@PRIMARYATIAS]

Νιcε ❤