Proving $\lim_{x\to\infty}xa^{x}=0$ in Elementary Ways

Question

I wish to prove the following limit without using L'Hopital rule or other known limits: $$\lim_{x\to\infty}xa^{x}=0$$ where $0<a<1$.

I wanted to do so using this sequence limit (which I know how to prove): $$\lim_{n\to\infty}na^{n}=0$$

I would appreciate to know if the following argument is valid (this is only the essence of it):

Let's denote for each $x>1$: $n_x=\lfloor x\rfloor$. We thus have for all $x>1$: $$0\le xa^{x}\le (n_{x}+1)a^{x}\le (n_{x}+1)a^{n_x}=n_{x}a^{n_x}+a^{n_x}$$

We can now use the fact that if $\lim_{k\to\infty}x_k=\infty$, than $\lim_{k\to\infty}n_{x_k}=\infty$, the limits $\lim_{n\to\infty}a^n,\lim_{n\to\infty}na^n=0$ and the squeeze theorem to get the desired result.

Of course we are using here properties of real exponents, which is ok for this discussion.

I would appreciate any feedback regarding this argument's validity.

Thanks a lot in advance!!

Answer 1

I think your proof sketch is solid.

An alternative route could be using elementary calculus (if that's elementary enough for you) to show that $xa^x$ is monotonically decreasing for $x>-\frac1{\ln a}$. Which means that for any $x$ larger than $1-\frac1{\ln a}$, we have $0<xa^x\leq n_xa^{n_x}$, again letting the squeeze theorem do the rest of the job.

Answer 2

Here's another approach that I think is easy to understand. Since the constant $a$ lies between $0$ and $1$, we can rewrite the term $a^x$ as the fraction $\frac{1}{c^x}$, where $c$ is a constant greater than $1$. This allows us to rewrite the limit as $\lim_{x\to\infty}\frac{x}{c^x}$. Since the denominator is an exponential term, while the numerator is a linear term, we can infer that since $c$ is greater than $1$, the denominator's value will eventually surpass the numerator's value as $x$ approaches infinity. This means the function value will approach $0$.

Answer 3

Another way could be as follows. We have $a^{-1}=1 +\delta.$ Then for $x>2,$ in view of the binomial identity, we get $$(1+\delta)^x\ge (1+\delta)^{n_x}\ge {n_x\choose 2 }\delta^2\ge {(x-1)(x-2)\over 2}\delta^2$$ Hence $$0<xa^x=x(1+\delta)^{-x}\le 2\delta^{-2}{x\over (x-1)(x-2)}\\ =2\delta^{-2}\left [{2\over x-2}-{1\over x-1}\right ]\to 0$$