How does one solve these kinds of nasty limits ?

Can you stop roasting me ,and show me how to do it without lhopital rule ,cause the book I'm studying from didn't mention that yet and it hit me with that limit

>l'hopital

That is cheating though..
What calculus course even allows that shit?

The limit is definitely less than 1.

en.wikipedia.org/wiki/E_(mathematical_constant)
has it being a consequence of n!~sqrt{2\pi n}(n/e)^n

You learn it in calc 1 first year so I don't see why you can't use it after that.

How the hell do you differentiate a factorial?

>ctrl+f "stirling"
>get nothing
use the goddarn stirling inequality mate
find it using some basic rectangle inequality or whatever you call it with the logarithm

>capital L for limit
triggered

First thing to do is to always (and I mean *always*) put everything in exponential form.
Whenever you see an exponent that depend on n, you write the whole expression in exponential form. Otherwise, you are guaranteed to be spewing bullshit (as there are no fixed rules for these situations).

L'hôpital rule tells you that if both f and g are defined near a and g != 0 at g(a), a being a real or +/-infinity, then the limit when x -> a of f/g is f'/g'.

So basically you can derivate that shit and work it out.

Fucking kill yourself brainlet

Gave it another try and reached a dead end here

Try to derivate it.
en.wikipedia.org/wiki/L'Hôpital's_rule

Have you tried writing n! as [math] n^n -sum_{i=1}^n(i) [/math]

there should be an n infront of the sum.

This is the answer

Work with the log of this, then use riemann sum

And the first part, that separation is perfectly valid

daily reminder that anybody who said L'Hopital is a fucking brainlet

One reads Rudin's Principles of Mathematical Analysis

en.wikipedia.org/wiki/Stirling's_approximation

you cant differentiate a factorial.

Yes you can, retard, factorial is the same as gamma fn, only defined on positive integers

> says you can
> explains why you can't

Lhopitals rule would work, as it is in indeterminate form, so long as you took the derivative of the factorial by approximating the gamma function to estimate n factorial.

What a nigger

Possible hint: If one takes the logarithm, one gets a Riemann sum for log.

What the fuck is this bullshit expression?

By differentiating the gamma function. It requires some analysis. Its also a silly thing to do as the factorial is defined as a function over the natural numbers.

>provable theorem
>cheating
undergrads actually think this

maa.org/sites/default/files/0002989019147.di991715.99p14142.pdf

L'hoptials rule will always give the correct answer if the limit is of type
[math]\frac{\infty}{\infty} \ \text{or} \ \frac{0}{0}[/math]

However if it is of different type; make sure that f and g are real functions that are differentiable on the open interval [math]a,b \in (-\infty,\infty)[/math]

In addition, make sure that following conditions are met aswell: [math]\mathrm{D}g(x) \neq 0[/math] for all [math] x \in (a,b) [/math] and that

[math] g(x) \to +\infty[/math] or [math]g(x) \to -\infty[/math] or [math] f(x) \to 0 \wedge g(x) \to 0 \ \text{when} \ x \to a^+ [/math]

When those conditions are met, it's only then you can apply l'hopitals rule and be sure it gives you a correct answer.

You can also solve it by doing some dodgy mathematics.

Let [math]a_n = \frac{\sqrt[n]{n!}}{n}=\left(\frac{n!}{n^n} \right)^{\frac{1}{n}} [/math]. IF the limit [math] \lim_{n\to\infty}a_n[/math] exists and is equal to L then we can do this.
Notice that
[math] \left( \frac{a_n}{a_{n+1}}\right)^n = \left(\frac{n+1}{n}\right)^na_{n+1} = \left(1+\frac{1}{n}\right)^n a_{n+1}[/math]
Take the limit to infinity of both sides then
[math] \left( \frac{L}{L}\right)^n = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n L [/math]
therefore
[math] 1 = e L [/math] and [math]L=\frac{1}{e}[/math] since [math]\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e[/math]

it shouldn't be (L/L)^n since n is the limit variable, my mistake but dodgy.

Use this equivalence n!=n^(n)*e^(-n)*√[2πn]
This only works when n goes to infinity.

nice equivalence.
I got another one for you
e^n = sqrt(n)

that seems pretty wonky.
At least from what I did (might not be exactly what you were suggesting)
You'd have to prove first, that the sum converges to the integral, since the partion size stays constant.
am I missing something there?

just figured it out i guess.
writing the sum like that takes away the wonkiness of the riemann integral

my first guess would be to apply a logarithm and see what happens

[math]\ln(n+1)[/math] is increasing.

So the upper Riemann sum is less than or equal to the integral.
[math]
\displaystyle
\begin{align}
\sum_{k=1}^n \ln(k+1) &\le \int_0^n \ln(t+1) dt \\
\ln(n!) &\le \ln(n+1) n - n \\
\frac{\ln(n!)}{n} - \ln(n) &\le \ln(1+\frac{1}{n}) - 1 \\
&\le -1, \, n \to \infty
\end{align}
[/math]

And the lower Riemann sum is greater than or equal to the integral.
[math]
\displaystyle
\begin{align}
\sum_{k=0}^{n-1} \ln(k+1) &\ge \int_0^n \ln(t+1) dt \\
\ln(n!) &\ge \ln(n+1) (n+1) - n \\
\frac{\ln(n!)}{n} - \ln(n) &\ge \ln(1+\frac{1}{n}) + \frac{\ln(n+1)}{n} - 1 \\
&\ge -1, \, n \to \infty
\end{align}
[/math]

Putting both of the above together.
[math]
\displaystyle
\begin{align}
\frac{\ln(n!)}{n} - \ln(n) &= -1, \, n \to \infty \\
\frac{\sqrt[n]{n!}}{n} &= \frac{1}{e}, \, n \to \infty
\end{align}
[/math]