Finding $\lim_\limits{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$ analytically 
$$\lim_{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$$

I can solve this easily by generating the graph with my calculator, but is there is a way to do this analytically?


I thought it might be instructive to present an approach that relies on elementary tools only.  To that end, we begin with the following primer.
PRIMER:  BOUNDS FOR THE LOGARITHM FUNCTION
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{x-1}\le \log(x)\le x-1} \tag 1$$
for $x<1$.


Now, we note that $\log_{1.1}(x)=\frac{\log(x)}{\log(1.1)}$.  In addition, we note that for any number $a$, we have $\log(x^a)=a\log(x)$.
Using $(1)$, we have for $a>0$,
$$\frac{x^{-1}-x^{-(a+1)}}{a}\le \frac{\log(x)}{x}\le \frac{x^{a-1}-x^{-1}}{a} \tag 2$$
Surely, $(2)$ is valid for $0<a<1$.  Then, applying the squeeze theorem to $(2)$ with $0<a<1$ (e.g., $a=1/2$) yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\frac{\log_{1.1}(x)}{x}=0}$$

and we are done!
