# Proving $\lim\limits_{n \to \infty} \frac{n^a}{c^n} = 0$ 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?

• hint: let $k$ be an integer such as $k>a$. Then $\frac{n^a}{c^n} < \frac{n^k}{c^n}$. Now apply L'Hopital $k$ times. – Vasya Aug 9 at 20:12
• Double-check that $1/c$ --- you're differentiating with respect to $n$, so $\ln c$ is a constant. – Neal Aug 9 at 20:18

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.

Your second differentiation is wrong because you've tried differentiating the denominator with respect to $$c$$ instead of $$n$$. If we differentiate $$a$$ times for an integer $$a\ge0$$, our limit is $$\lim_{n\to\infty}\frac{a!}{c^n\ln^a c}$$, which $$=0$$ for $$c>1$$. (You can handle other values of $$a$$ by squeezing.)

Hint If $$c>1$$, the limit is trivial for $$a \leq 0$$. For $$a>0$$ show instead that $$\left( \lim\limits_{n \to \infty} \frac{n^a}{c^n} \right)^\frac{1}{a}=0$$

Then, raise both powers to $$a$$.