I'm looking at another interesting thing in Desmos, if you assume n is any real number: for any b and c >0, the limits when n->0+ and n->0- are, respectively, a and 1 if a>=1, and 1 and a otherwise. It's continuous at 0 only if a=1. Just plug this in: y=\left(\frac{a^{\frac{1}{x}}+b}{c} ight)^{x}
You may replace 1/n by h, and after this transformation use Hospital rule. In this form you differentiate a sequence. You obviously know it works, but in formal you are wrong. To avoid this type of problems you may use Stolz theorem
Question about case 2. As c > b+1, couldn't we just say [ (a^(1/n) + b) / c ]^n < [ (a^(1/n) + b) / (b + 1) ]^n and then apply the limit on both sides? By case 1, the right hand side converges to a^(1/(b+1)), so the limit in question is bounded by a finite number and thus itself finite. What am I missing?
Let s(x) be the sum of all divisors of x including itself, for examle s(7)=8, s(10)=18, let's create a sum of 1/s(x) as x goes from 1 to infinity. Is that convergent and where does it converge to?
I think there's something wrong with the proof in case 2, because you start by saying that (nthroot(a)+b)/c can get arbitrarily close to (b+1)/c as n gets greater and greater, but then derive the limit as 0 - but if the limit is 0, then clearly at some point as n gets greater and greater the expression starts getting further away from (b+1)/c in order to get closer to 0. Case 3 seems to have not made this mistake, so it's just a matter of copying that argument and modifying it for case 2.
As it's already said it depends on convention. If you consider just real numbers R , infinity is out of domain. For instance limit of 1/x² (x goes to zero) would not exist. If you consider extended real line with infinity, then it does exist and it is equal to infinity. HOWEVER , the limit of 1/x doesn't exist even if we consider extended real line , since the limit diverges to different infinities.
One of my pet peeves I have for my Calculus students, one cannot take a derivative of a function defined using a discrete variable. n is defined to be from the positive integers. You can't apply L'H as you can't apply the derivative. All that needed to be done was to state. Something needed to be addressed for this situation. Michael could have just said, "If we view n to be a real variable, and the limit of the function converges, then so does the function on the positive integers."
Technically speaking the problem didn’t say n is an integer. Maybe it’s often used to denote integers, but just because you use n to denote your variables, doesn’t necessarily mean by some unbreakable mathematical law that it has to be an integer.
@@divisix024 Technically you are correct; he can choose to use any variable he wants. Now considering all his other videos (at least the ones I have seen, and I have been following him for years) have n viewed as an integer. Contextually, one would assume it is an integer. Therein lie the problem. It is ambiguous. And ambiguity should not be the responsibility of the viewer. In the grand scheme of things, this is a tiny tiny tiny part of his whole video. Still, I see my students try to use L'H/derivatives on discrete functions (e.g. lim n->∞ (-1)^n n^2/3^n does not work with L'H if we view n as a real or continuous variable). He could have avoided the ambiguity by providing a single statement of context.
Either I'm half asleep or he completely botched taking the derivative of the numerator for l'Hopital's rule in case 1. The value is a composition f(g(n)) where f(x) is ln(x) and g(x) is (a^(1/x)+b) [after expressing the n'th root as the 1/n'th power] (d/dx)(f(g(x)) is f'(g(x))g'(x) so the derivative should be (1/(a^(1/n)+b))(1/n)(a^(1/n-1)) This simplifies to (a^(1/n-1))/(n(a^(1/n)+b)) whose limit as n approaches infinity is zero. So ln(L)=0 and L=1. Or, as I just mentioned, maybe I'm half asleep.
No because you used the power rule instead of the derivative of the exponential in a^(1/n). Remember here n is the variable and a is a constant so we have another chain rule here a^(1/n) = e^(ln(a)/n) (a^(1/n))' = e^(ln(a)/n) . (ln(a)/n)' = a^(1/n) . ln(a) . -1/n²
