If the limit exists, you can take the limit into the exponent because f(x)=e^x is a continuous function. First show the limit exists and then use the continuity of e^x.
Precise, clear, well presented, well decomposed +elegance and + well spoken english (I'm not a native english teacher). You are definetly one of my favourite math youtuber ! :)
...Good evening Newton, I hope your doing well. You look like a "mathematical" alchemist, who manages to tackle every seemingly unsolvable limit! Now, only the easiest thing left for you, namely to create gold out of anything! (lol). Another great performance with visibly a lot of enthusiasm. Thank you Newton, Jan-W
Well cant you rewright this as 1÷x^1÷x and by ising l hopital rule,you will the limit as x apriches infinity of 0÷((-x^1÷x-2)(ln(x)-1))=0÷(-x^1÷x)ln(x)×x² and bc x^1÷x aproches 1 and bc x² grows much faster than ln(x) than that expresion aproches 0 so you will get 0÷0 and in limit standards it equals 1
6:20 there's another proof that we use in class that is the limit of a function/x tell us its parabolic branch direction , if it follows the vertical axis it'll be infinity, and if it follows the horizontal axis it'll be 0
Lol I did it by intuition and I was correct. My way of thinking was that we have something raised to the power of something that is close to 0 and therefore the answer is 1. But number e can also come in handy here. Imho you overcomplicated a bit with introducing logarithm and y, it can be solved faster If you know the formula for e: (1+1/x)^x when x tends infinity is e. Just transform initial equation to this and you will essentially get the same result.
In algebra, 0^0 is defined and = 1. analysis workers don't have the same opinion. as we use limits here, we not define it. However, (1/x)^(1/x) -> 1 still. (0^0 been controversial for 2 centuries and still is) to me it's as much defined as 3x1/3 = 1 and not 0,99999999999999999... Edit : tho what happens when x < 0 ? -normally not defined in analysis because of ln- ? (ie. (1/-0.5)^(1/-0.5) => -2^(-2) => 1/(-2)² = 1/4 thus, is defined.
Not all limits are taught before derivatives. Some limits require more sophisticated strategies. That's where L'Hopital's rule comes in. There also other strategies.
8 minutes to solve an elementary exercise that can be solved in few seconds! Changing variable y = 1/x the limit is reduced to the limit as y goes to 0^+ of y^y, that is exp(ylog y). Now ylog y goes to 0 and so y^y goes to 1.
If you have to do it by observation anyway after doing all that math, then why can’t you do the same in first step, 1/x to power of 1/x is zero to power of zero which is equal to 1?
My logic was a bit more dumb. I just said , ok , its like 1/(x^(1/x)) this is continuous , it can't do to many jums, so lets just find n^(1/n). But this is much more elegant
Только у вас постояннпя ошибка , когда вы пишитп логарифм, с именно основание не указано. Log имеет любое основание хоть 2, хоть 10, хоть е . Поэтому десятичный логарифм имеет запись Lg= log(10)
The base approaches ∞ and therefore you have the indeterminate form ∞^0. That's why you just can't calculate the limit when the function is written as x^(-1/x).
@diegocabrales since the exponent gets smaller and smaller, the fact that it's negative is less and less significant, meaning the base can be considered approaching infinity
@@jamesharmon4994You're right that the base doesn't approach 0 but ∞-my mistake. I've already edited my comment. But that just only changes the indeterminate form: now we have ∞^0 one. The thing is that the result is different depending on which are the paths that both functions take to approach 0 and ∞, so you just can't say that ∞^0 = 1 because it could be convergent to another number, or diverge to ∞, or -∞, or even not be defined.
@@jamesharmon4994 Yes, you're right, n^0 = 1 for every number n ≠ 0. If n = 0 then we have 0^0, which is undefined. But ∞ is not a number, so you can't apply it to ∞. ∞ is a limit approach and ∞^0 is an indeterminate from, meaning it could be any number, or diverge to ∞, or -∞, or even not be defined. All depends on how do our functions approach ∞ and 0, respectively. That's why you just can't say that ∞^0 = 1 and be ok with that. Let me use another indeterminate form to understand it. You could say that since all real powers of 1 are equal to 1, then 1^∞ = 1. However, let me introduce you to e, Euler's number: e = lim(x --> 0)(1 + x)^(1/x) Clearly the base approaches 1, and 1/x approaches ∞. However, 1^∞ = e ≠ 1. But we can have 1^∞ = 1, like in this example: lim(x --> ∞)(1 + 1/2^x)^x = 1 And we can continue doing that with two functions which make the indeterminate form 1^∞ and be equal to any number, or diverge to ∞, or -∞, or even not be defined. Then what you apply to numbers can't be applied in general to limits.
The limit of that function is zero. and if you check the numbers you can use, you need to know: X is an element of the real numbers, but without zero. if you look left and right from zero, it goes to plus infinity.
In my very humble opinion, the proof here is wrong. The limit at infinity of 1/X to the power of 1/x is 0 to the power of 0 (as the limit at infinity of 1/X is 0). But 0^0 has no definite agreed value (only non zero number to the power of 0 are equal to 1). I suspect that replacing the limit of the function by the function of the limit is a trick that does not work if the limit does not exist. So still in my humble opinion, the proof works because you furst define and accept that 0^0 is defined.
@@geraldvaughn8403My mistake. I meant that he accepts that x^0=1. He said that 0^0 is not 1 but e^0=1. First to be clear, I know what official academia says. It's a convention that x^0=1 but it never stood with me because the exponent definition is clear and it says that a^b means "take 0 and add a multiplied with itself b times". Now don't get confused. Actually multiplication happens b-1 times but you get the point. So, if b=0 what would be the logical result? 0 or 1?
