Question

The value of $\underset{x\to \infty }{lim}\frac{\ln x-\left[x\right]}{\left[x\right]-\left\{x\right\}}$ is
where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part.

Answer 1

$L=\underset{x\to \infty }{lim}\frac{\ln x-\left[x\right]}{\left[x\right]-\left\{x\right\}}$
$=\underset{x\to \infty }{lim}\frac{\ln x-\left[x\right]}{\left[x\right]-\left\{x\right\}}$
$=\underset{x\to \infty }{lim}\frac{\frac{\ln x}{\left[x\right]}-1}{1-\frac{\left\{x\right\}}{\left[x\right]}}$
$=\underset{x\to \infty }{lim}\frac{\ln x}{\left[x\right]}-1$

Now, $\underset{x\to \infty }{lim}\frac{\ln x}{\left[x\right]}$ is $\left(\frac{\infty }{\infty }\right)$ form
$=\underset{x\to \infty }{lim}\frac{\ln x}{x}$
$=\underset{x\to \infty }{lim}\frac{1/x}{1}$
$=\underset{x\to \infty }{lim}\frac{1}{x}=0$

$\therefore L=0-1=-1$