limit of x-ln(x) as x goes to infinity via L'Hospital's Rule

Рет қаралды 26,258

2 жыл бұрын

limit of x-ln(x) as x goes to infinity via L'Hospital's Rule. This is an indeterminate form of infinity - infinity so we must "do more work"! If you know my secret weapon, The List, then you can say that x will be so much bigger than ln(x) as x goes to infinity, so much bigger so that x-ln(x) is infinity. But this calculus 1 tutorial shows you why it is true.
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Пікірлер: 53

@owoLight 2 жыл бұрын

nice solution, what i did was put the inside to the power with base e, and then ln on the outside of the limit. with exponent properties you’ll have e^x/e^ln(x) in the limit which simplifies to e^x/x. Use L’H and then evaluate from there!

@OriginalSuschi 2 жыл бұрын

I did the same

@mauriciocaviedes6520 6 ай бұрын

Thanks pal!! It was very helpful! 😊

@The1RandomFool 2 жыл бұрын

Alternatively, x=log(e^x). You can then combine logarithms and move the limit inside the logarithm.

@justabunga1 6 ай бұрын

An alternative way to do that instead of factoring out x is to look at x as ln(e^x). By using properties of logarithms ln(e^x)-ln(x)=ln(e^x/x). l'Hopital's rule is used since it's in an indeterminate form of ∞/∞. We must take the derivative of top and derivative of bottom separately. This means that the ln of (limit as x goes to ∞ of e^x/1 or e^x) is limit as x goes to ∞ of x. The limit is therefore ∞.

@eugeneimbangyorteza 2 жыл бұрын

Even without solving, it's obvious that for very large numbers, x overtakes ln(x)

@mengasketo253 2 жыл бұрын

there is actually a inequality proof that says lnx

@Eagle-yc9nr Жыл бұрын

yeah it may be obvious but you gotta show your work otherwise no credit is given

@browhat6935 2 жыл бұрын

Just speedrunned this entire playlist, thanks a lot, it really helped!

@amirmahdypayrovi9316 2 жыл бұрын

Similar question to practice: question 3 HMMT2004: hmmt-archive.s3.amazonaws.com/tournaments/2004/feb/calc/problems.pdf Answer(BY L"H): 2/3

@tr060 2 жыл бұрын

What about lim ln (e^x) - ln x = lim ln ( e^x / x) Then l'Hopital etc etc...?

@davidw6936 2 жыл бұрын

That’s what I was thinking too.

@Pankaw 2 жыл бұрын

That was the first thing I thought

@Pankaw 2 жыл бұрын

x grows faster and faster than ln(x) as x approaches infinity (since d/dx(ln(x)) = 1/x approaches zero, where as d/dx(x) = 1 stays constant), so it's clear that x - ln(x) will approach infinity and ln(x)/x will approach 0

@Apollo23682 7 ай бұрын

Thanks! This finally makes sense

@darius2915 2 ай бұрын

Great channel!

@francoislaveaux7072 2 жыл бұрын

Where did you get your Shirt ?

@star_ms 2 жыл бұрын

Can you do a video on integral of 1/x wrtx as x goes from -∞ to ∞? Good video by the way!

@serafimbarbu7711 Жыл бұрын

Thank you very much

@heliocentric1756 2 жыл бұрын

*In general*: If f/g approaches infinity as x approaches a, then: limit of f-g = limit of f, as x approaches a. (a can be finite or infinite)

@ffc1a28c7 2 жыл бұрын

No doubt about it, but do ya have a proof?

@heliocentric1756 2 жыл бұрын

@@ffc1a28c7 f-g = f(1-g/f) Since f/g approaches infinity as x approaches a, so g/f approaches 0. Thus f-g and f have the same limit as x approaches a.

@NoNameAtAll2 2 жыл бұрын

dear just_calculus, please add all the other your channels into Channels tab on this channel too, not only on bprp!

@tonicastanares5270 2 жыл бұрын

What happens if instead of going for 1-0 right away, you distribute the infinity to both 1 and 0. Wouldn't that make the second one indeterminate?

@RicharD-fi3ho 5 ай бұрын

Thanks

@zombieslayer2225 2 жыл бұрын

I was wondering if this is allowed: after factoring out x, I used limit laws to break up the limit. limit x--------> infinity (x) ( Iimit x--------> infinity (1) - limit x--------> infinity (lnx/x) ) It gave me infinity (1-0) which is infinity.

@JonathanMandrake 2 жыл бұрын

I would use the exponential function, giving us e^x /x, using L'Hospital this is the same as e^x /1=e^x and thus infinity, and since e^x=infinity, x=infinity also

@iGasteiz 2 жыл бұрын

Super!

@ucansusamuru594 4 ай бұрын

how can I find this t-shirt !!!!!

@paulocoelho301 Жыл бұрын

and if it was lim x--> 0 so in the same equation, would change anything?

@snipergranola6359 2 жыл бұрын

I did just opposite take logx common

@darekdarek6233 2 жыл бұрын

The proposed solution is incorrect because it leads to an indefinite (indeterminate) symbol 0 * infinity (you have it on your T-shirt anyway). You should represent the expression x- ln (x) as one logarithm, and then examine the argument boundary of this resulting logarithm. It is exactly the limit of ln (e ^ x / x). It is enough to examine the boundary of the argument e ^ x / x. Applying the L'Hospital rule we investigate the limit e ^ x which is infinity (because the derivative of the numerator to e ^ x and denominator is 1). The limit of the natural logarithm of the argument that goes to infinity is infinity. This is an unequivocal solution.

@zoeconstantinidou4303 8 күн бұрын

Greattt

@BenjaminFaltin 2 жыл бұрын

Without using L'H we could say that x - lnx = ln(e^x/x), then with Taylor's expension for e^x we just get 0 + a bunch of infinities

@poubellestrange7515 2 жыл бұрын

Here is a solution without L'Hôpital: L = lim (x - ln x) e^L = lim e^(x - ln x) = lim (e^x)/(e^ln x) = lim (e^x)/x Let y = e - 1. Therefore, by the Binomial Theorem, for x > 3, e^x = (y + 1)^x ≥ 1 + yx + y²(x - 1)x/2 + ... > y²(x - 1)x/2 We have e^L = lim (e^x)/x ≥ lim (y²(x - 1)x)/(2x) = (y²/2)( lim (x - 1) ) = ∞ Since e^L = ∞, this implies that L = ∞.

@user-sw4id2bs5r 2 жыл бұрын

何言ってるかは分からないけど、何やってるかだけは分かるの最高に数学してる。

@moorsyjam 2 жыл бұрын

If you multiply x - ln(x) by (x + ln(x))/(x + ln(x)), you end up with (x² - ln⁡(x)²)/(x + ln⁡(x)), which is of the infty/infty form

@nicholasroberts2933 2 жыл бұрын

its not actually, the numerator is infty - infty which is indeterminate, so L'Hopitals cannot be applied

@moorsyjam 2 жыл бұрын

@@nicholasroberts2933 Ah, of course it is. I split the fraction into x²/(x + ln⁡(x)) - ln⁡(x)²/(x + ln⁡(x)), used L'H, and completely missed that.

@ezecattalin8 2 жыл бұрын

Isn't it still indeterminate because (1-0) isn't exactly 1 but approaching 1?

@arvamin5896 2 жыл бұрын

But (lnx)' = 1/x

@user-liusteven302 2 жыл бұрын

老師好像看得懂中文？我就用中文留言了在把lnx/x“計算”成0那步應該不對，因為它不是因子，不能先將重要極限算出來有誤歡迎指正

@hi-dg1bx 8 ай бұрын

可以吧先用arithmetic rule分别evaluate两个limit limit of 1 exist evaluate lnx/x 发现是0， limit也exist 外面的x当成是另外一个function乘进去同样也是arithmetic rules的东西单纯是x 趋向于无限的话也不是indeterminate form，所以♾️x 1 就是无限