If $0< a < 1$; show that $\lim na^n$ goes to $0$ as $n$ goes to $\infty$
I know that when $0< a < 1$, $\lim a^n$ goes to $0$ and $\lim n$ goes to $\infty$
$|a^n| <\epsilon$ take $\epsilon_0 = \epsilon/|n|$ then $|a^n| < \epsilon_0 = \epsilon/|n|$, then $|n||a^n|< \epsilon$, and since $|na^n|\le|n||a^n|< \epsilon$, then $|na^n|< \epsilon$
Is this kind of argument sufficient for this case?
Here a very elementary way:
So, it follows $$0 \leq na^n = \frac{n}{(1+p)^n} < \frac{n}{\color{blue}{\frac{n(n-1)}{2}p^2}} = \frac{2}{(n-1)p^2} \stackrel{n \to \infty}{\longrightarrow} 0$$
When treating sequences of reals, one does not deal strictly speaking with L'Hopital's formula (which applies to derivable functions) but with its discrete counterpart, namely the Cesaro-Stolz criterion. Set $b=\frac{1}{a}>1$; in your particular case, you can apply this criterion to the sequence of ratios: $$\left(\frac{n}{b^n}\right)_{n \in \mathbb{N}}$$
which reduces finding the limit of this said sequence to that of the sequence $$\frac{1}{b^{n+1}-b^n}=\frac{1}{b^{n}(b-1)} \xrightarrow {n \to \infty} 0$$
assuming of course you know what the asymptotic behaviour of the exponential sequence is.
One can also consider a different argument based on an inequality both powerful and very fundamental to Analysis. In it's most general form it can be stated as:
Generalised Bernoulli inequality: for $a \in (0, \infty)\setminus \{1\}$ and $x, y \in \mathbb{R}^{*},\ x<y$ it holds that $$\frac{a^x-1}{x}<\frac{a^y-1}{y}$$
We are going to focus on a particular version of this, easily established by induction, which claims that:
Simplified version: for $a>1$ and $n \in \mathbb{N}, n \geqslant 2$ one has $$a^n>1+n(a-1)$$
This can be used in a crafty way to derive: $$a^n=(\sqrt[k]{a}\ ^{n})^k>(1+n(\sqrt[k]{a}-1))^k>n^k(\sqrt[k]{a}-1)^{k}$$
for arbitrary $k \in \mathbb{N}^*$. Can you see how this entails that an exponential of strictly superunitary (strictly above $1$) base asymptotically dominates powers?
Another option is to note that $xa^x=x\exp(-|\ln a|x)$ is a positive function for $x>0$, reaching a maximum at $x=1/|\ln a|$, and $\int_0^\infty x\exp(-|\ln a|x)dx=1/\ln^2 a$ is finite. This requires the $x\to\infty$ limit to be $0$.
Let's make use of the general theorem that if $a_n\to L$, then ${a_1+\cdots+a_n\over n}\to L$, where $a_k\in\mathbb{R}$ for all $k$ and $L\in\mathbb{R}\cup\{\pm\infty\}$ (so that sequences that "converge" to infinity are encompassed by the theorem).
If $A\gt1$, then we have
$$\lim_{n\to\infty}{A^n\over n}=\lim_{n\to\infty}{A^n-1\over n}={A-1\over A}\lim_{n\to\infty}{A+A^2+A^3+\cdots+A^n\over n}={A-1\over A}\lim A^n=\infty$$
It follows that if $0\lt a\lt1$, then $1/a\gt1$, so that
$$\lim_{n\to\infty}na^n={1\over\lim_{n\to\infty}{(1/a)^n\over n}}={1\over\infty}=0$$
We can use the l’Hospital.
$$\lim_{n\to \infty} n\cdot a^n=\lim_{n\to \infty} \dfrac{n}{a^{-n}}$$ I think we can agree on the fact that the nominator and denominator converge to infinity. Therefor
$$\lim_{n\to \infty} \dfrac{n}{a^{-n}}= \lim_{n\to \infty} \dfrac{\dfrac{d}{dn}n}{\dfrac{d}{dn}a^{-n}}=\lim_{n\to \infty}\dfrac{1}{-a^{-n}\cdot \log(a)} = 0$$
Sińce $\ln n \leq 0.25n$ we have $$0.5 n\ln a \geq \ln n + n\ln |a| $$ For sufficiently large $n.$ Therefore for $n\geq\frac{2\ln\varepsilon}{\ln a}$inequality holds.