How to prove $\lim_{n\to{\infty}}\frac{x^n}{n!} = 0$  My try:  I know that factorial function increases much more rapidly than exponential function but how to really prove it.
1447
Why is     $$\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}$$   Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is     $$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$     This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .
119655
Determine when $\sum_1^\infty \frac{(2n)!x^n}{n(n!)^2}$ converges. By ratio test, when $|x|&lt;1/4$, the sum converges, $|x|&gt;1/4$ diverges. But I'm not sure about $|x|=1/4$. By Stirling's approximation $\frac{(2n)!(1/4)^n}{n(n!)^2}\sim\frac{4^n}{n{4^n}\sqrt{ \pi n}}\sim\frac{1}{\sqrt{\pi}n^{3/2}}$, so when $|x|=1/4$, the sum converges. But I haven't learned Stirling's approximation yet, so how to prove that when $|x|=1/4,$ the sum converges, without using it?
52332
I have a question which is giving me a hard time.    I want to show that $$ \lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x$$ given that $\lim_{n\to \infty} x_n= x$.
38578
Prove that:  $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0$$  I think I need to show $2^n \geq n^3$ $\forall n \geq10$. Is it true?  What about: $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0$$  What to do with $n!$?
82582
$\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$ is equal to: a) $e$ b) $2e$ c) $\sqrt e$ d) $e^2$  Though it looks really innocent at first sight, it certainly isn't. Attempt: It's $\infty^{\infty}$ form. I had tried taking the product raised to the power $\frac{1}{n(n+1)}$ as function $f(n)$. Then I took logarithm of both sides to see if things simplifying.  Even after factoring out the extra factorials it wasn't easy. Note that ${^n\mathrm{C}_x} = \binom{n}{x}$.
119777
How can I show $\lim_{x \to \infty} \frac{1}{x} = 0$ using epsilon delta proof.  Its pretty obvious that the limit is zero, but I am still new at epsilon proofs.
68123
I was trying to work out a problem I found online. Here is the problem statement:     Let $f(x)$ be continuously differentiable on $(0, \infty)$ and suppose $\lim\limits_{x \to \infty} f&#39;(x) = 0$. Prove that $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$.    (source: )  The first idea that came to my mind was to show that for all $\epsilon &gt; 0$, we have $|f(x)| &lt; \epsilon|x|$ for sufficiently large $x$. (And I believe I could do this using the fact that $f&#39;(x) \to 0$ as $x \to \infty$.)  However, I was wondering if there was a different (and nicer or cleverer) way. Here's an idea I had in mind:  If $f$ is bounded, then $\frac{f(x)}{x}$ clearly goes to zero. If $\lim\limits_{x \to \infty} f(x)$ is either $+\infty$ or $-\infty$, then we can apply l'Hôpital's rule (to get $\lim\limits_{x \to \infty} \frac{f(x)}{x} = \lim\limits_{x \to \infty} \frac{f'(x)}{1} = 0$).   However, I'm not sure what I could do in the remaining case (when $f$ is unbounded but oscillates like crazy). Is there a way to finish the proof from here?   Also, are there other ways of proving the given statement?
110771
I am trying to show that $\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0$. I found that using stirling's approximation, I can get: $$ \lim_{n\to \infty}{\frac{2n\choose n}{4^n}}= \lim_{n\to \infty}{\frac{(2n)!}{n!^2*2^{2n}}}= \lim_{n\to \infty}{\frac{(\frac{2n}{e})^{2n}\sqrt{2\pi (2n)}}{((\frac{n}{e})^n\sqrt{2\pi n})^2*2^{2n}}} =\lim_{n\to \infty}{\frac{2\sqrt{\pi n}}{2\pi n}} =0 $$  But this seems inelegant where there should be a more elegant, combinatorial proof. Is there an easier way?
57939
$$\lim_{n \to \infty }n\int_{0}^{1}\frac{x^{n}}{1+x+x^{n}}dx$$  I factorize $x^n$ then I tried with $u=1+x^{1-n}$ but I didn't get too far. Also I tried to get something from $0&lt;x&lt;1$. I know that if $x$ is from $(0,1)$ then $x^n$ tends to $0$ as $n \to \infty$.
47213
$\lim n^n/(e^nn!)=0$ using Stirling approximation it is obvious. But can we do it without using Stirling approximation.  Now series with terms $x^n n^n/n!$ has ROC $1/e$. What we can say about behaviour of power series at endpoints?
10877
How can I prove $\lim_{n \to \infty} \frac{n^2}{2^n}=0$  I tried to use $\mid \frac{n^2}{2^n} - 0 \mid &lt;\epsilon $ However, because of $n^2$ I cannot use it. Also, I tried to use ratio test and I got $\lim_{n-&gt;\infty}\frac{(n+1)^2}{2}\frac{1}{n^2}$, but after that one, I dont know how to get limit &lt; 0  I have no idea to solve this problem.   I can also use a squeez lemma, but dont know how to apply to this question.
8977
$$\lim_{n \to\infty} \frac{n!}{n^n}$$  I have a question: is it valid to use Stirling's Formula to prove convergence of the sequence?
65752
I need to find the limit for:   $ \lim_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n} $  I know the answer is $\frac {1}{e}$ but I have no idea how to get that answer. I'd appreciate some help.
78897
$\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$  I let, $x = \frac{1}{n}$, then as   $\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$  L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$  Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.
86379
I have to solve: $\lim\limits_{x\to \infty} \frac{\ln(x+1)}{\ln(x)}$. Can you give me any hints to go? Thanks a lot!
32524
How can I find the $\lim  (1+(a^{1/n}-1) / b)^n$ when $n \to \infty$?    I would greatly appreciate it if you kindly give me some hints.
52366
How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$  without using L'Hospital or Taylor series?   thanks :)