Finding the limit of a function?

Question

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.

There seems to be a habit on this site of referring to Guillaume François Antoine, Marquis de l'Hôpital as Mr Hospital. — Tim, May 18 '13 at 09:08

score 2 · Answer 1 · answered May 18 '13 at 09:25

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 $$ \frac{t^{k+1}}{e^t} < \frac{t^{k+1}}{t^{k+2}} = \frac{1}{t} \xrightarrow{t \to \infty} 0. $$

score 1 · Answer 2 · answered May 18 '13 at 09:24

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.

score 1 · Accepted Answer · answered May 18 '13 at 09:34

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.

score 1 · Answer 4 · answered May 18 '13 at 10:09

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. $$\lim_{t\to+\infty}\frac{t^{k}}{e^t}=0.$$

Finding the limit of a function?

4 Answers4