How to prove $\lim_{n\rightarrow\infty}nx^n=0$ without L'Hôpital's rule, where $x \in [0,1)$?? How to prove $$\lim_{n\rightarrow\infty}nx^n=0$$ without L'Hôpital's rule? where $x \in [0,1)$ and $n=1,2,3,...$.
I know one of way to prove this is to treat $n$ is real, and $n$ and $(\frac{1}{x})^n$ as functions of $n$. Then, apply L'Hôpital's rule and it works for discrete $n$ by the definiation of limit. 
However, I would like to skip the step we treat $n$ is real. Is there any proof for this question only using the techniques for limit of sequences? (such as the sandwich theorem, etc.)
 A: Let $x<1-1/N$.  If $n>2N$ then $(n+1)x^{n+1}/(nx^n)<(1+\frac1{2N})(1-\frac1N)<1-\frac1{2N}$.  So terms from then on shrink by at least that factor, and $nx^n<C(1-\frac1{2N})^n\to0$.
A: Hint: When $|x| < 1$ we have that $$\sum_{n=1}^{\infty} n x^n = \frac{x}{(x-1)^2}$$
A: Let $0\lt x\lt 1$. Then there is a positive $a$ such that $x=\frac{1}{1+a}$. By the Binomial Theorem, we have for $n\ge 2$ that 
$$0\lt (1+t)^n =1+nt+\frac{n(n-1)}{2}t^2\gt \frac{n(n-1)}{2}t^2.$$
It follows that
$$0\lt nx^{n} \lt \frac{2}{t^2}\cdot \frac{1}{n-1}.$$
Now it is straightforward, given a positive $\epsilon$, to produce an $N$ such that $0\lt nx^n \lt \epsilon$ if $n\gt N$.
A: For $0 < x < 1$ we have $x = (1+y)^{-1}$ with $y > 0$.
It follows from the binomial theorem that,
$$ (1+y)^n > \frac{n(n-1)y^2}{2}.$$
Hence,
$$nx^n = \frac{n}{(1+y)^n}< \frac{2}{(n-1)y^2}.$$
For any $\epsilon > 0$, if $n > 1 + 2/(\epsilon y^2)$, then $|nx^n-0|= nx^n < \epsilon$.
Thus, $\lim_{n \to \infty} nx^n = 0$.
A: Let $x \in ]0, 1[$ (the case where $x = 0$ is trivial). Since
$$
\frac{(n+1)x^{n+1}}{nx^{n}} = (1+\frac{1}{n})x \to x < 1,
$$
the series $\sum_{n} nx^{n}$ converges, so $nx^{n} \to 0$.
I hope this is helpful.
A: If $x\lt1$, then find $n_x$ so that $\frac1{n_x}\le\frac{1-x}2$. Then, for any $k\ge n_x$, $\frac1k\le\frac{1-x}2\implies x\le\frac{k-2}k$ and
$$
\begin{align}
\frac{(k+1)x^{k+1}}{kx^k}
&=\frac{k+1}kx\\
&\le\frac{k+1}k\frac{k-2}k\\
&=1-\frac1k-\frac2{k^2}\\
&\le\frac{k-1}k
\end{align}
$$
Thus, for any $n\gt n_x$,
$$
\begin{align}
\frac{nx^n}{n_xx^{n_x}}
&=\prod_{k=n_x}^{n-1}\frac{(k+1)x^{k+1}}{kx^k}\\
&\le\prod_{k=n_x}^{n-1}\frac{k-1}k\\
&=\frac{n_x-1}{n-1}
\end{align}
$$
Therefore,
$$
\begin{align}
\lim_{n\to\infty}nx^n
&\le\lim_{n\to\infty}\frac{n_xx^{n_x}(n_x-1)}{n-1}\\[6pt]
&=0
\end{align}
$$
where the numerator is a fixed constant depending on $x\lt1$.
