Question

If $\left\{x\right\}$ denotes the fractional part of $x$, then
$\underset{x\to 0}{lim}\frac{\left\{x\right\}}{tan\left\{x\right\}}=$

• A. 1
• B. 0
• C. -1
• D. none of these

Answer 1

The correct option is D none of these

We have,
$\underset{x\to 0^{-}}{lim}\frac{\left\{x\right\}}{tan\left\{x\right\}}$
$=\underset{x\to 0^{-}}{lim}\frac{x-\left[x\right]}{tan(x-[x\left]\right)}=\underset{x\to 0^{-}}{lim}\frac{x+1}{tan(x+1)}=\frac{1}{tan1}=cot1$ and,
$\underset{x\to 0^{+}}{lim}\frac{\left\{x\right\}}{tan\left\{x\right\}}=\underset{x\to 0^{+}}{lim}\frac{x-\left[x\right]}{tan(x-[x\left]\right)}=\underset{x\to 0^{+}}{lim}\frac{x}{tanx}=1$
Clearly, $\underset{x\to 0^{-}}{lim}\frac{\left\{x\right\}}{tan\left\{x\right\}}\ne \underset{x\to 0^{+}}{lim}\frac{\left\{x\right\}}{tan\left\{x\right\}}$
So, $\underset{x\to 0}{lim}\frac{\left\{x\right\}}{tan\left\{x\right\}}$ does not exist.