Is the following limit true or not? (function convergence)

Question

Let $x\in(0,1)$, let $p>1.$ I was wondering if $\lim_{n\to\infty} n^{\frac{2}{p}}x^{\frac{n}{p}}=0$ is true or not? I plotted some values and it seems to be true but I am struggling to prove this. Intuitively, one thing I can think of is that $x^{\frac{n}{p}}$ decays much quicker than $n^{\frac{2}{p}}$.

Answer 1

Your intuition

that $x^{\frac{n}{p}}$ decays much quicker than $n^{\frac{2}{p}}$.

is basically correct. In general, exponential values of the form $y^z$ where $y \gt 1$, with $z \to \infty$, always eventually grow faster than any polynomials $p(z)$.

Since $x\in(0,1)$, then

$$y = \frac{1}{x} \gt 1 \tag{1}\label{eq1A}$$

As such, you can rewrite your limit as follows

$$\begin{equation}\begin{aligned} \lim_{n\to\infty} n^{\frac{2}{p}}x^{\frac{n}{p}} & = \lim_{n\to\infty}\frac{n^{\frac{2}{p}}}{y^{\frac{n}{p}}} \\ & = \lim_{n\to\infty}\left(\frac{n^{2}}{y^{n}}\right)^{\frac{1}{p}} \\ & = \left(\lim_{n\to\infty}\frac{n^2}{y^{n}}\right)^{\frac{1}{p}} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Since you have the limits are $\infty$ in both the numerator & denominator, you can use L'Hôpital's rule. Using it twice gives

$$\lim_{n\to\infty}\frac{n^2}{y^{n}} = \lim_{n\to\infty}\frac{2}{(\ln(y))^2 y^{n}} = 0 \tag{3}\label{eq3A}$$

This shows the limit in \eqref{eq2A} is also $0$.

Answer 2

\begin{align*} \lim_{n \rightarrow \infty} n^{\frac{2}{p}}x^{\frac{n}{p}} &= \lim_{n \rightarrow \infty} \mathrm{e}^{\frac{2}{p}\ln n}\mathrm{e}^{\frac{n}{p}\ln x} \\ &= \lim_{n \rightarrow \infty} \mathrm{e}^{\frac{2 \ln n + n \ln x}{p}} \end{align*}

Since $0<x<1$, $\ln x < 0$, so $n \ln x$ is a function linear in $n$ with negative slope. $2 \ln n$ is logarithmic in $n$. As you observe, the linear function dominates, so

\begin{align*} \lim_{n \rightarrow \infty} \mathrm{e}^{\frac{2 \ln n + n \ln x}{p}} &= \lim_{n \rightarrow \infty} \mathrm{e}^{\frac{n \ln x}{p}} \\ &= \lim_{n \rightarrow \infty} x^{\frac{n }{p}} \\ &= 0 \text{.} \end{align*}