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

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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#justcalculus

[owoLight]

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!

[justabunga]

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]

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

[TheRandomFool]

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

[browhat]

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

[tr]

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

[Pankaw]

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

[heliocentric]

*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)

[paulocoelho]

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

[zombieslayer]

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

[NoNameAtAll]

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

[tonicastanares]

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?

[poubellestrange]

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 = ∞.

[darekdarek]

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.

[JonathanMandrake]

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

[BenjaminFaltin]

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

[ezecattalin]

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

[moorsyjam]

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

[star_ms]

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

[teelo]

So the equation "I hate you times infinity plus one" really does equal "doh".