Question

Let $\underset{x\to 0}{lim}\frac{[x{]}^2}{x^2}=l$ and $\underset{x\to 0}{lim}\frac{\left[x^2\right]}{x^2}=m$, where $[.]$ denotes greatest integer, then:

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

Answer 1

Answer: (B)

$m$ exists but $l$ does not

${lim}_{x\to 0^{+}}\frac{[x{]}^2}{x^2}={lim}_{h\to 0}\frac{0^2}{h^2}=0$
${lim}_{x\to 0^{-}}\frac{[x{]}^2}{x^2}={lim}_{h\to 0}\frac{(-1{)}^2}{(-h{)}^2}=\infty$
$l$ does not exists at $x=0$
${lim}_{x\to 0^{+}}\frac{\left[x^2\right]}{x^2}={lim}_{h\to 0}\frac{\left[h^2\right]}{h^2}=0$
${lim}_{x\to 0^{-}}\frac{\left[x^2\right]}{x^2}={lim}_{h\to 0}\frac{\left[\right(-h{)}^2]}{(-h{)}^2}=0$
$m$ exists at $x=0$