How to solve the limit as $n\to \infty$ of $n^2\cdot x^n$ when $x$ is bounded to (0,1)?

Question

So because $x$ is between 0 and 1, limit of $x^n$ is $0$ and limit of $n^2$ is $\infty$ so it's indeterminate.

I can't figure out how to solve it. I would like to not use l'Hopital's rule when solving this.

Answer 1

Consider writing $$ a_n = n^2 \cdot x^n = e^{\log(n^2)} \cdot e^{\log(x^n)} = \exp( 2 \log(n) + n \log(x) ). $$ Recall that $\exp$ is a continuous, strictly increasing function. So it suffices to consider the limit of the exponent. Note that $n \gg \log n$ when $n$ is sufficiently large. Now think about the sign of $\log(x)$ for $x \in (0,1)$. This can give you the limit of the exponent, so now you just need to convert that into the limit of the original expression.

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For this final step, we see that $\log(x) < 0$ for all $x \in (0,1)$, and so $$ b_n = 2 \log(n) + n \log(x) \to -\infty \quad \text{as $n \to \infty$}. $$ This means that for all $K > 0$ there exists an $n_0$ (depending on $K$) such that $$ b_n \le -K \quad \forall \ n \ge n_0. $$ This, in turn, tells us that $$ a_n = \exp(b_n) \le e^{-K} \quad \forall \ n \ge n_0,$$ since $\exp(\cdot)$ is an increasing function.

Now, clearly $a_n \ge 0$ for all $n$. Set $\epsilon = e^{-K}$, ie $K = \log(1/\epsilon)$, and we see the above is equivalent to the following: $$ \forall \epsilon > 0 \ \exists n_0 \text{ s.t. } 0 \le a_n \le \epsilon \ \forall n \ge n_0. $$ This precisely says that $a_n \to 0$ as $n \to \infty$.

Answer 2

If you can use series, then the ratio test for $\sum n^2 x^2$ gives that that series converges when $x \in (0,1)$ and so $n^2 x^2 \to 0$.

Answer 3

Note that if I take the binomial expansion of $$1^{n+3}=\left(x+(1-x)\right)^{n+3}= x^{n+3}+ \dots +\frac {(n+3)(n+2)(n+1)}{6}x^n(1-x)^3+\dots$$ I have (all terms are positive) $$1\gt\frac {(n+3)(n+2)(n+1)}{6}x^n(1-x)^3\gt \frac {n^3}6x^n(1-x)^3$$ so that $$\frac 6{n(1-x)^3}\gt n^2x^n$$

You should be able to conclude from there.

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This proof, picking out just one term from the binomial expansion, looks to be too wasteful to be truly effective, but it should be obvious how to adapt it to $n^rx^n$ for an arbitrary positive integer $r$. It runs surprisingly smoothly and doesn't use sophisticated machinery, so worth noting, I think, even if other methods are preferred. This kind of trick appears elsewhere too.