How to prove that $\lim_{n \to \infty} n x^{n} = 0 $ when $0Intuitively it's easy, but hard to prove by the epsilon-delta method:
$$ \lim_{n \to \infty} n x^{n} = 0$$
 A: We have $nx^n=\exp(\log n)\exp(n\log x)=\exp(n\log x+\log n)$, so it's enough to show that $n\log x+\log n\to -\infty$ as $n\to +\infty$. We use the fact that $\log n\leq \sqrt n$ for $n$ large enough to see that 
$$n\log x+\log n\leq n\log x+\sqrt n=n\left(\log x+\frac 1{\sqrt n}\right).$$
As $\log x<0$, $\log x+\frac 1{\sqrt n}<\frac{\log x}2$ for $n$ large enough hence 
$$n\log x+\log n\leq n\frac{\log x}2,$$
which gives the result.
A: Let $a_n=nx^n$. You have
$$\frac{a_{n+1}}{a_n}= \frac{n+1}n x = \left(1+\frac1n\right) x.$$
Let $q=\frac{1+x}2$. (Or we could choose any $q$ such that $x<q<1$.)
There exists $n_0$ such that $\left(1+\frac1n\right) x \le q$ for $n\ge n_0$. (Since $\lim\limits_{n\to\infty} \left(1+\frac1n\right) x = x < q$.)
Thus we have 
$$0 \le a_n \le a_{n_0} \cdot q^{n-n_0}$$
for $n\ge n_0$. (You can show this easily by induction.)
Since $q^n\to 0$ for $n\to\infty$ we get
$$\lim_{n\to\infty} a_n=0.$$
A: Suppose you know $\lim n^{\frac{1}{n}} =1$, let $\epsilon >0$
then take $n_o$ such that 
$$n^{\frac{1}{n}} \leq 1+\frac{a-x}{x} \forall n \geq n_0$$ where
$$x<a<1$$
Now assume you know $\lim b^n =0$ for $b\in (0,1)$
Choose a $n_1$ such that $$a^n \leq \epsilon \forall n \geq n_1$$
Now take $ \max\{n_0,n_1\}$
so we get
$$nx^n= (n^{\frac{1}{n}}x)^n \leq  ((1+\frac{a-x}{x})x)^n=a^n\leq \epsilon$$
A: I would propose another approach. Notice that the sequence $$b_n:=\frac{1}{x^n}$$ is positive, strictly increasing and unbounded. 
Define $$a_n:=n$$ as well.
Stolz Cesaro theorem is applicable and therefore, if $$\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ exists and it is equal to $l\in\mathbb R$, then also the limit proposed is equal to $l$. 
Then we are reduced to evaluate $$\lim_{n\to\infty}\frac{1}{\frac{1}{x^{n+1}}-\frac{1}{x^n}}=\lim_{n\to\infty}\frac{x^{n+1}}{1-x}=0.$$ To tackle the last equality the $\varepsilon$-$\delta$ argument is perfectly fine and easy to perform. To conclude, this argument finishes the proof. 
In a way you have not cheated too much because $\varepsilon$-$\delta$ is used. Hope you liked it. Mathematicians are lazy :P.
Bye 
A: Well, here again I'll try a fancy proof. Let us look at the power series
$$\sum_{n=1}^\infty nx^n\,\,,\,\,\text{and let us define}\,\,\,a_n:=nx^n$$
We try the ratio test to find this series convergence radius:
$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(n+1)x^{n+1}}{nx^n}\right|=|x|\frac{n+1}{n}\xrightarrow [n\to\infty]{} |x|$$
Thus, the series converges (absolutely, even) for $\,|x|<1\,$ , from which it follows that the series general term must converge to zero, i.e. 
$$a_n=nx^n\xrightarrow [n\to\infty]{} 0\,\,,\,\,\text{for}\;\;|x|<1$$
A: Isn't it enough to know that the exponential function with $a>1$ goes faster to infinity than the polynomial function? 
$$\lim_{n\to\infty}\frac{n}{a^n}=\lim_{n\to\infty}\frac{e^{\ln n}}{e^{n \ln a}}=0.$$
Chris. 
A: This will follow from a more general result.
 If $\{a_n\}\geq 0$ be a monotone decreasing sequence such that $\sum a_n <\infty$ then $na_n\rightarrow 0.  $
Let  $\epsilon >0$ be arbitrary . Then there exists a natural number $m$ such that for $p\in \mathbb N $
$$a_{m+1}+a_{m+2}+..+a_{m+p}<\epsilon $$ But $\{a_n\}$ is monotone decreasing means 
$$pa_{m+p}\leq a_{m+1}+a_{m+2}+..+a_{m+p }< \epsilon$$ putting p-=m we get 
$$2ma_{2m}< 2\epsilon$$ Also putting $p=m+1$ we get
$$(m+1)a_{2m+1}<\epsilon$$ Hence 
$$(2m+1)a_{2m+1}\leq (2m+2)a_{2m+1}<2\epsilon$$
Hence both the sequences $\{2ma_{2m}\}$ and $
\{(2m+1)a_{2m+1}\}$ converges to $0$. Hence the result follows. 
A: We prove the result under the slightly weaker condition $|x|\lt 1$.
Let $|x|=\dfrac{1}{1+t}$. Then $t\gt 0$. 
By the Binomial Theorem, if $n \ge 2$, then 
$$(1+t)^n \ge 1+nt +\frac{n(n-1)}{2}t^2 \gt \frac{n(n-1)}{2}t^2.$$
It follows that 
$$0\le n|x^n| \lt \frac{2}{(n-1)t^2}.$$
Now it is easy, given $\epsilon \gt 0$, to find $N$ such that if $n \gt N$ then 
$\dfrac{2}{(n-1)t^2}\lt \epsilon$.
Remark: If we do not wish to use the Binomial Theorem, let $m=\left\lfloor\frac{n}{2}\right\rfloor$. By the Bernoulli Inequality, $(1+t)^m \ge 1+mt$, and therefore $(1+t)^n \ge (1+mt)^2\ge m^2 t^2$. 
A: I hope this works like that. It sounds too simple.
Define $y = \frac{1}{x}$ We know $0 \lt x \lt 1$ that means $y \gt 1$.
$$ \lim_{n\to\infty} (nx^n) = \lim_{n\to\infty} (\frac{n}{y^n}) $$
We have now “$\frac{\infty}{\infty}$“ because $y \gt 1$ and can use L‘Hopitals Rule.
$$ \lim_{n\to\infty} (nx^n) = \lim_{n\to\infty} (\frac{n}{y^n}) = \lim_{n\to\infty} \frac{1}{ny^{n-1}} = 0$$
