$\lim 2^n r^{n^2}$ if $0<r<1$?

Question

I'm trying to find for some $0<r<1$ $$\lim_{n\rightarrow\infty}2^nr^{n^2}.$$ I plugged in a few numbers and know that it should go to zero but I don't know how to justify it using limits.

Answer 1

Note that

$$L := \lim_{n \to \infty} 2^n r^{n^2} \implies \ln(L) = \lim_{n \to \infty} n \ln(2) + n^2 \ln(r)$$

Clearly, the latter term will outpace the former in terms of growth, so

$$\ln(L) = \lim_{n \to \infty} n \ln(2) + n^2 \ln(r) = \ln(r) \lim_{n \to \infty} n^2 = -\infty$$

since $r \in (0,1) \implies \ln(r) < 0$. Thus, since $L \ge 0$ clearly, we can conclude,

$$L = \lim_{x \to -\infty} e^x = 0$$