calculus - Sequence convergence with inequality

Find the limits:

$$lim_{n \to \infty} \frac{n^4 -n^3}{e^n}$$

My Attempt:
using L'Hospital's rule I got that the limit equals to $0$.

(1) the sequence $a_n$ given by:
$$a_n = \int_{1}^{t} e^{-nx} x^3 dx, \ n=1,2,\ldots$$

then $a_n$ is the function value $F(n)$, as:

$$F(t) = \int_{1}^{t} e^{-nx} x^3 dx$$

(2) use the mean value theorem on the function $F(t)$ in the interval $[1,n]$ to show that: $$\frac{a_n}{n-1} = e^{-nc} c^3, \ c\in (1, n)$$

My Attempt:
I think I'm able to show this. Is it correct that $f(b)-f(a) = f(b) = a_n$ because $f(a) = o$ (integral from 1 to 1)?

(3) why is $$e^{-nc} c^3 (n-1) \le e^{-n} n^{3} (n-1), c\in (1,n)$$

My Attempt:
Need help, I don't know how to approach this. What is the role of $c$ here?

(4) Use the inequality to show that the sequence $a_n$ converges, and find the limit: $lim_{n\to \infty} a_n$

My Attempt:

(4) Not sure what to do here, maybe I have to solve (3) first.

Answer

For (3), it's very simple:

• $c>1$ implies $-nc<-n$, and as the exponential is an increasing function, $\mathrm e^{-nc}<\mathrm e^{-n}$.


• the $x^3$ function is also increasing, so $ c implies $c^3

As these inequalities involve only positive numbers, we can multiply them to get
$$e^{-nc}c^3
and there only remains to multiply both sides by the positive number $n-1$.