Finding the limit of a function? The limit is actually easy: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t}$
One can use hopitals rule and say that ultimately the upper function will be reduced to a constant while the lower function will remain the same. Hence, the limit is zero. I was wondering if there is another way or a more mathematical representation of the answer. 
 A: You can also use the series representation $e^t = \sum_{n=0}^\infty \frac{t^n}{n!}$ to observe that for any given $k \in \mathbb{N}$ we have $e^t > t^{k+2}$ for $t > 1$. Hence
$$
\frac{t^{k+1}}{e^t} < \frac{t^{k+1}}{t^{k+2}} = \frac{1}{t} \xrightarrow{t \to \infty} 0.
$$
A: You could notice that 
$$\begin{array}{rcl} \lim_{t\to\infty} \frac{t^{k+1}}{e^t} &=& \lim_{t\to\infty} \frac{((k+1)t)^{k+1}}{e^{(k+1)t}}\\ &=& (k+1)^{k+1}\lim_{t\to\infty} \left(\frac t{e^t}\right)^{k+1} \end{array}$$
Now you can use continuity of $x\mapsto x^{k+1}$ and you only need to use L'Hôpital once.
A: Alternatively, we can proceed by induction on $k\in \mathbb{N}$.
Base Case: For $k=0$, we need only apply L'Hôpital's rule once: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t}{e^t} = \lim \limits_{t\to\infty}\dfrac{1}{e^t} = 0$.
Inductive Hypothesis: Assume that $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t} =0$ holds for $k=r$.
It remains to prove that the equation holds for $k=r+1$. Then by one more application of L'Hôpital's rule and the induction hypothesis, observe that:
$$\lim_{t\to\infty}{\frac{t^{r+2}}{e^t}}=\lim_{t\to\infty}{\frac{(r+2)t^{r+1}}{e^t}}=(r+2)\lim_{t\to\infty}{\frac{t^{r+1}}{e^t}}=(r+2)(0)=0$$
as desired.
A: Without l'Hôpital:
We start with  $$\tag1e^x\ge 1+x\quad\text{for all }x\ge 0,$$
 which follows (and in fact holds even for all $x\in\mathbb R$)


*

*from Bernoulli's inequality applied to $\left(1+\frac xn\right)^n$ if one introduces $e^x$ as $\lim_{n\to\infty}\left(1+\frac xn\right)^n$

*from $e^x\ge 1$ for $x\ge0$ and $e^x-e^0=(x-0)e^\xi\ge x$ with $0<\xi< x$ (mean value theorem) if one introduces $e^x$ as solution of $y'=y$ with $y(0)=1$

*by dropping all higher powers if one introduces $e^x$ via the power series $\sum_{k=0}^\infty\frac{x^k}{k!}$


By substituting $x=\frac t{k+1}$ in $(1)$ and raising to $(k+1)$st power, this becomes
$$ e^t=\left(e^{t/(k+1)}\right)^{k+1}\ge \left(1+\frac t{k+1}\right)^{k+1}\ge \frac1{(k+1)^{k+1}}\cdot t^{k+1}\quad\text{for }t\ge 0.$$
Therefore,
$$\left|\frac{t^{k+1}}{e^t}\right|\le (k+1)^{k+1}\quad\text{for }t\ge 0$$
and
$$\left|\frac{t^{k}}{e^t}\right|\le (k+1)^{k+1}\cdot\frac1t\quad\text{for }t> 0,$$
i.e.
$$\lim_{t\to+\infty}\frac{t^{k}}{e^t}=0.$$
