Question

$\underset{x\to 0}{lim}{x}^5\left[\frac{1}{x^3}\right]$ where [.] denotes the greatest integer function.

• A. 0
• B. 1
• C.
• D. -1

Answer 1

The correct option is A

0

We have, $\underset{x\to 0^{+}}{lim}{x}^5\left[\frac{1}{x^3}\right]$

We know that $x-1<\left[x\right]\le x$ $xϵR$

so, $\frac{1}{x^3}$ - 1 < $\left[\frac{1}{x^3}\right]$ $\le$ $\frac{1}{x^3}$

When x > 0

$\Rightarrow$ $x^5$ $(\frac{1}{x^3}-1)$ < $x^5$ $\left[\frac{1}{x^3}\right]$ $\le$ $\frac{x^5}{x^3}$ for x > 0

Taking Limit $\underset{x\to 0^{+}}{lim}$ throughout the inequality.

$\underset{x\to 0^{+}}{lim}$ $(x^2-x^5)$ < $\underset{x\to 0^{+}}{lim}$ $x^5$ $\left[\frac{1}{x^3}\right]$ $\le$ $\underset{x\to 0^{+}}{lim}{x}^2$

$0\le \underset{x\to 0^{+}}{lim}{x}^5\left[\frac{1}{x^3}\right]\le 0$

By sandwich theorem $\underset{x\to 0^{+}}{lim}{x}^5\left[\frac{1}{x^3}\right]=0$

Also, when x < 0

$\frac{1}{x^3}$ - 1 < $\left[\frac{1}{x^3}\right]$ $\le$ $\frac{1}{x^3}$

Multiplying $x^5$ in the equation

Inequality sign changes

$\frac{x^5}{x^3}\le x^5\left[\frac{1}{x^3}\right]\le x^5(\frac{1}{x^3}-1)$

Taking $\underset{x\to 0^{-}}{lim}$ through out the inequality