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?
 A: Since the limit of both the top and bottom is $\infty$ alone, l'hopital's rule gives us
\begin{align*}
\lim_{x \to \infty}\frac{\log_{1.1} x}{x}
&= \lim_{x \to \infty}\frac{\ln x}{(\ln 1.1) x} \\
&= \lim_{x \to \infty}\frac{(1/x)}{(\ln 1.1)} \\
&= 0.
\end{align*}
A: hint: L'hospitale's rule ! . Have you learned this formula yet?
A: Let $g(x)=\ln(x)-\sqrt{x}$
we have $g(1)=0$ and
$$g'(x)=\frac{1}{2x}(2-\sqrt{x})$$
$g$ is decreasing at $[4,+\infty)$
thus
$$\forall x\geq 4  \;\;0<\frac{\log_{1.1}(x)}{x}\leq \frac{1}{\ln(1.1)\sqrt{x}}$$
and squeeze theorem.
A: Use the definition: $\;\log_{1.1}x=\dfrac{\ln x}{\ln 1.1}$, so
$$\frac{\log_{1.1}x}{x}=\frac 1{\ln 1.1}\dfrac{\ln x}{x}\xrightarrow[x\to+\infty]{}\frac 1{\ln 1.1}\cdot 0=0.$$
A: 
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!
