I have a litle discution with a friend about the folowing limit:
$$\lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n}$$
I would solve it like this: $$\lim_{n\to\infty} \sqrt[n]{\frac{n!}{n^{n}}} =0$$
or
$$\lim_{n\to\infty} \frac{\sqrt[n]{n}\sqrt[n]{n-1}\cdots\sqrt[n]{1}}{n}=\frac{1*1*1\cdots}{\infty}=0$$
and in this 2ยบ way would there be a problem with $1^{\infty}$? I would say that no, because there is no functions involved, since as much as I know this undetermination is because you would whant to avoid the situation such as $f(x)^{g(x)}$ where $f(x)\to1$ and $g(x)\to\infty$ Could anyone clarify this for me?
Saturday, 21 February 2015
real analysis - limit of nth root of factorial devided by n
Subscribe to:
Post Comments (Atom)
real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
$$ 3x+6y+5z=7 $$ The general solution to this linear Diophantine equation is as described here (Page 7-8) is: $$ x = 5k+2l+14 $$ $$ y = -l $...
-
How to show the following inequality in Measure Theory: If $f$ is a non-negative measurable function defined on a measurable set $E$ then ...
-
I need help to compute the following integral: $$\int_{-\infty}^{\infty}\frac{z^4}{1+z^8}dz$$ I need to use Cauchy's residue theorem. I ...
No comments:
Post a Comment