A calculus problem by crazy singh

this limit changes into

$a= \lim _{ x\to \infty } \left( n\frac { \ln { { x } } }{ \left[ x \right] } -1 \right)$

now we need to calculate $\lim _{ x\to \infty } \left( \frac { \ln { { x } } }{ \left[ x \right] } \right)$

we know that $x-1\quad \le \quad \left[ x \right] \quad \le \quad x$

so $\frac { 1 }{ x-1 } \quad \ge \quad \frac { 1 }{ \left[ x \right] } \quad \ge \quad \frac { 1 }{ x }$

and $\frac { \ln { x } }{ x-1 } \quad \ge \quad \frac { \ln { x } }{ \left[ x \right] } \quad \ge \quad \frac { \ln { x } }{ x } ;\quad$ since $\ln { x }$ is positive as $x$ tends to positive infinity.

$\lim _{ x\to \infty } \frac { \ln { x } }{ x-1 } \ge \lim _{ x\to \infty } \frac { \ln { x } }{ \left[ x \right] } \ge \lim _{ x\to \infty } \frac { \ln { x } }{ x }$

$0\quad \ge \lim _{ x\to \infty } \frac { \ln { x } }{ \left[ x \right] } \ge \quad 0$ ;

since $\lim _{ x\to \infty } \frac { \ln { x } }{ x-1 }$ and $\lim _{ x\to \infty } \frac { \ln { x } }{ x }$ both are equal to zero. (can be proved by L' Hospital's rule, $\frac { \infty }{ \infty }$ form )

hence $\lim _{ x\to \infty } \frac { \ln { x } }{ \left[ x \right] } = 0$

so $a=(n\times 0)-1=-1$ and ${ a }^{ 2 }=1$