Evaluate $$L=\lim_{x \to \infty} \frac{ \ln \lfloor x \rfloor}{\lfloor x \rfloor}$$
My try:
I tried using
$$x-1 \lt \lfloor x \rfloor \le x $$ $\implies$
$$ \frac{1}{x} \le \frac{1}{\lfloor x \rfloor} \lt \frac{1}{x-1}$$
$\implies$
$$\frac{\ln \lfloor x \rfloor }{x} \le \frac{ \ln \lfloor x \rfloor}{\lfloor x \rfloor} \lt \frac{ \ln \lfloor x \rfloor}{x-1} \tag{1}$$
Now $$\lim_{ x \to \infty} \frac{\ln \lfloor x \rfloor }{x}=\lim_{ x \to \infty} \frac{ \ln x+\ln \left(1-\frac{\left\{x\right\}}{x}\right)}{x}$$
$$\lim_{ x \to \infty} \ln \left(1-\frac{\left\{x\right\}}{x}\right)=0$$
Since $$\lim_{ x \to \infty} \frac{\left\{x \right\}}{x}=0$$
Also $$\lim_{x \to \infty}\frac{\ln x}{x}=0$$
Hence $$\lim _{ x \to \infty}\frac{\ln \lfloor x \rfloor }{x}=0$$
Hence By Squeez theorem in $(1)$ we get $$L=0$$
Is this the right approach?