Question

Let $\underset{x\to 0^{+}}{lim}\frac{[x{]}^2}{x^2}=l$ and $\underset{x\to 0^{-}}{lim}\frac{[x{]}^2}{x^2}=m$, where $[.]$ denotes greatest integer. Then

• A. $l$ exists but $m$ does not
• B. $m$ exists but $l$ does not
• C. both $l$ and $m$ exists
• D. neither $l$ nor $m$ exists

Answer 1

The correct option is A $l$ exists but $m$ does not

As $x\to 0^{+}\Rightarrow \left[x\right]=0$
$\therefore l=\underset{x\to 0^{+}}{lim}\frac{[x{]}^2}{x^2}=\underset{x\to 0^{+}}{lim}\frac{0}{x^2}=\frac{0}{\text{A very small non zero number}}=0$
Now
$x\to 0^{-}\Rightarrow \left[x\right]=-1$
$\therefore m=\underset{x\to 0^{-}}{lim}\frac{[x{]}^2}{x^2}=\underset{x\to 0^{-}}{lim}\frac{-1}{x^2}=-\infty$
Hence $l$ exist but $m$ does not exist.