calculus - Proving $limlimits_{n to infty} frac{n^a}{c^n} = 0$ using L'Hôpital's Rule

I am trying to prove $\displaystyle \lim_{n \to \infty} \frac{n^a}{c^n} = 0$ using L'Hôpital's Rule, but I'm stuck.

Here's what I have so far:

$$\lim_{n \to \infty} \frac{n^a}{c^n} = \lim_{n \to \infty}\frac{an^{n-1}}{c^n \ln c} = \lim_{n \to \infty}\frac{a(a-1)n^{a-2}}{c^n(\ln c)^2 + c^n \frac{1}{c}}$$

All three limits above seem to evaluate to $\frac{\infty}{\infty}$, so I feel like I'm not getting anywhere. Any ideas?

Edit: So, with the help of the hints below, I was able to figure out that

$$\lim_{n \to \infty} \frac{n^a}{c^n} = \frac{a}{\ln c} \cdot \lim_{n \to \infty} \frac{n^{a-1}}{c^n} = \frac{a}{\ln c} \cdot \frac{a - 1}{\ln c} \cdot \lim_{n \to \infty} \frac{n^{a-2}}{c^n} = \cdots$$

So, disregarding the constant, it looks like the numerator keeps decreasing, while the denominator stays the same.

I can also see that if I let $a = 2$, for instance, I end up with $0$ after applying L'Hopital's $2$ times:

$$\begin{aligned} \lim_{n \to \infty} \frac{n^2}{c^n} &\overset{LH}= \lim_{n \to \infty} \frac{2n}{c^n \ln c} \\ &= \frac{2}{\ln c} \lim_{n \to \infty} \frac{n}{c^n} \\&\overset{LH}= \frac{2}{\ln c} \lim_{n \to \infty} \frac{1}{c^n \ln c} \\ &= \frac{2}{(\ln c)^2} \lim_{n \to \infty} \frac{1}{c^n} \\ &= 0 \end{aligned}$$

So it seems reasonable to conclude that for an arbitrary $a > 0$, I will end up with $0$ after applying L'Hopital's $a$ times.

But I'm not sure how to go about using induction to prove it formally. I've only proven very simple sums by induction so far. Do I have to apply it to a product here?

Answer

I deleted my old answer, as it missed the point a bit (especially given the edits to the question). I'm going to expand on J.G.'s answer, since you seem to need a little extra help.

Let's prove $\lim_{n\to\infty} \frac{n^a}{c^n} = 0$, for $a \in \Bbb{R}$ and $c > 1$. (if $0 < c \le 1$, then the sequence does not tend to $0$, and for $c = 0$, the expression is undefined). We can tackle this in a number of cases, but the cases reduce back down to one case fairly easily, using the squeeze theorem.

Case 1: $a \in \Bbb{N}_0 = \{0, 1, 2, \ldots\}$, and $c > 1$
In this case, we use induction on $a$ (not $n$, as I originally suggested). When $a = 0$, then

$$\frac{n^a}{c^n} = \frac{1}{c^n}.$$
This tends to $0$, a fact which you seem happy to assume. If you wished to prove it, observe that the sequence $a_n = \frac{1}{c^n}$ satisfies is decreasing, bounded below by $0$, and hence convergent. It also satisfies the recurrence relation $a_{n+1} = \frac{a_n}{c}$, so if $L$ is its limit, then taking the limit of both sides yields $L = \frac{L}{c} \implies (c - 1)L = 0$, and hence $L = 0$, as $c \neq 1$.

You probably could skip the above proof, but either way, the base case is established.

Now, suppose for some $k \in \Bbb{N}_0$ (and $c > 1$), we have

$$\lim_{n \to \infty} \frac{n^k}{c^n} = 0.$$
Then,
$$\begin{align*} \lim_{n \to \infty} \frac{n^{k+1}}{c^n} &= \lim_{n \to \infty} \frac{(k+1)n^k}{\ln c \cdot c^n} &\text{L'Hopital's rule} \\ &= \frac{k+1}{\ln c} \lim_{n \to \infty} \frac{n^k}{c^n} \\ &= \frac{k+1}{\ln c} \cdot 0 = 0 &\text{induction hypothesis.} \end{align*}$$

By induction, we now have $\lim_{n \to \infty} \frac{n^a}{c^n} = 0$ for all $a \in \Bbb{N}_0$ and $c > 1$. That is, we have completed this case.

Case 2: $a \in \Bbb{R}$, and $c > 1$
To prove this case, simply choose any natural number $k$ such that $k \ge a$ (we can do this, due to the Archimedean property). Naturally, if we take a negative value of $a$, then just choose $k = 0$ (or $1$, or anything higher really). Then, note that for all $n$,

$$0 \le \frac{n^a}{c^n} \le \frac{n^k}{c^n}.$$
The first case proved that $\frac{n^k}{c^n} \to 0$. Thus, by squeeze theorem, we have a proof for case 2.

We can even extend to $c < -1$ too!

Case 3: $a \in \Bbb{R}$, and $c < -1$
We prove this again by squeeze theorem. Note that,

$$-\frac{n^a}{|c|^n} \le 0 \le \frac{n^a}{|c|^n},$$

and by case 2, both bounds tend to $0$, proving case 3.

Hope that helps, and sorry for the misleading hint.