2. Re: limit

Originally Posted by dhiab
See here.

3. Re: limit

Originally Posted by dhiab
Also, dhiab, I see that you're a Super Member. Maybe this is at least partially a challenge. Can you address that?

Let $\displaystyle \ \ ln(X) \$ mean $\displaystyle \ \ log_e(X)$.

Let y = $\displaystyle \displaystyle\lim_{n \to \infty}(5n)^{1/n}$

ln(y) = ln[$\displaystyle \displaystyle\lim_{n \to \infty}(5n)^{1/n}]$

ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}ln[(5n)^{1/n}]$

ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\bigg(\dfrac{1}{n}\bigg)ln(5n)$

ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\dfrac{ln(5n)}{n}$

I'll use L'Hopital's Rule:

ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\dfrac{\bigg(\dfrac{1}{n}\bigg)}{1}$

ln(y) = 0

$\displaystyle e^{ln(y)} = e^0$

y = 1

The limit is 1.

4. Re: limit

Hello i'have this solution :

lim(n--> + infini) (1/n)Ln(5n) = 0 .

lim(n--> + infini) racine_nième(5n)

= lim(n--> + infini) (5n)^(1/n)

= lim(n--> + infini) exp(Ln((5n)^(1/n)))

= lim(n--> + infini) exp((1/n)Ln(5n)) = 1 .

5. Re: limit

Originally Posted by dhiab
Define $x_n=\sqrt[n]n-1$
If $n\ge 2$ then
\begin{align*}n&=(1+x_n)^n \\n&\ge\dfrac{n(n+1)}{2}(x_n)^2\\\text{or }(x_n)^2&\le\dfrac{2}{n+1}\\\text{as }n\to\infty\text{ then }(x_n)&\to 0\\\therefore \sqrt[n]n-1&\to 0\\\therefore \sqrt[n]n&\to 1 \end{align*}

6. Re: limit

Originally Posted by greg1313
I'll use L'Hopital's Rule:

ln(y) = $\displaystyle \displaystyle\lim_{n \to \infty}\dfrac{\bigg(\dfrac{1}{n}\bigg)}{1}$
I don't think much of your differentiation.

7. Re: limit

Originally Posted by Archie
I don't think much of your differentiation.
You don't think much what of my differentiation!? You made
a negative comment without a specific reason attached
to it. So, you're trolling, and you wasted a post. Learn to make

8. Re: limit

Keep your hair on. It was a joke - that rather backfired as my differentiation wasn't up to much, now I look at it again.

9. Re: limit

Originally Posted by Archie
Keep your hair on. It was a joke - that rather backfired as my differentiation wasn't up to much, now I look at it again.
You're an expert on all sorts of differentiation and integration techniques on multiple
message board sites.
Thank you.

10. Re: limit

Originally Posted by greg1313
You don't think much what of my differentiation!? You made
a negative comment without a specific reason attached to it. So, you're trolling, and you wasted a post.
TO: greg1313, O.K. I can understand your being insulted by that comment. But I will add that depending on the level of the course, I would not accepted a solution using differentiation, l'Hopital's Rule. In my analysis courses I always went for more general principles.
This was one of my favorite question. SEE my post above: $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n} = 1$
Having proved that, it an easy step to show that for $m\ge 0$ then $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{m} = 1$
Of course $\sqrt[n]{5n}=\sqrt[n]{5}\sqrt[n]{n}$. Now that is a real teaching moment, a general concept. Not a one time off problem.

11. Re: limit

Originally Posted by Plato
Define $x_n=\sqrt[n]n-1$
If $n\ge 2$ then
\begin{align*} . . . \therefore \sqrt[n]n&\to 1 \end{align*}
This is a start/promising. And we are still left to prove it with the
And suppose the radicand were $\displaystyle n^2, \ n^3, \ \ or \ \ even \ \ n^{ \sqrt{n}}$.
We would need to show that those limits are also 1, or not.

12. Re: limit

Originally Posted by greg1313
This is a start/promising. And we are still left to prove it with the
And suppose the radicand were $\displaystyle n^2, n^3, or even n^{ \sqrt{n}}$.
We would need to show that those limits are also 1, or not.
Surely this is not too much for you:
$\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{n^m}}} = {\left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n}} \right)^m} = {1^m} = 1~?$

13. Re: limit

Originally Posted by Plato
Surely this is not too much for you:
$\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{n^m}}} = {\left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n}} \right)^m} = {1^m} = 1~?$
That is a trap. If m = n (for instance), it collapses to the nth root of (n raised
to the n exponent). Then, it is the limit as n approaches infinity
of just n, which equals infinity.

14. Re: limit

Originally Posted by greg1313
That is a trap. If m = n (for instance), it collapses to the nth root of (n raised
to the n exponent). Then, it is the limit as n approaches infinity
of just n, which equals infinity.
You do not understand much about notation do you? In $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{n^m}}} = {\left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{n}} \right)^m}$ anyone claiming to help at this level surely realizes that $n$ is a variable but $m$ is constant. So given your limitation, please in the further consider if you are up to answering a question.

15. Re: limit

Originally Posted by Plato
You do not understand much about notation do you?
anyone claiming to help at this level surely realizes that $n$ is a variable but $m$ is constant. So given your limitation, please in the further consider if you are up to answering a question.
You do not understand much about not making presumptions, do you!? So, given your limitation in your lack of ability to make a valid discussion, please refrain from making them,
because you cross into making uncivil posts. Your history of it is clear. Stop doing it, hurry up and understand that. Your post was reported. And you haven't been learning from