Evaluate $$L=\lim_{x \to \infty} \frac{ \ln \lfloor x \rfloor}{\lfloor x \rfloor}$$

My try:

I tried using

$$x-1 \lt \lfloor x \rfloor \le x $$ $\implies$

$$ \frac{1}{x} \le \frac{1}{\lfloor x \rfloor} \lt \frac{1}{x-1}$$


$$\frac{\ln \lfloor x \rfloor }{x} \le \frac{ \ln \lfloor x \rfloor}{\lfloor x \rfloor} \lt \frac{ \ln \lfloor x \rfloor}{x-1} \tag{1}$$

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?

  • 3
    $\begingroup$ $$\lim_{x \to \infty} \frac{ \ln \lfloor x \rfloor}{\lfloor x \rfloor}=\lim_{n \to \infty} \frac{ \ln n}{n}$$ $\endgroup$ – Nosrati Jun 16 '18 at 12:42
  • 1
    $\begingroup$ Rather squeeze between $\frac{\ln(x-1)}x$ and $\frac{\ln x}{x-1}$, I propose $\endgroup$ – Hagen von Eitzen Jun 16 '18 at 13:09
  • 1
    $\begingroup$ The result in question is an immediate/obvious corollary of the standard limit $\lim_{x\to \infty} (\log x) /x=0$. There is no need for all this squeeze if you already know the standard limit. $\endgroup$ – Paramanand Singh Jun 17 '18 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.