Question

The value of $\underset{x\to 0}{lim}\frac{\left(\tan \left(\right\{x\}-1)\right)\sin \left\{x\right\}}{\left\{x\right\}\left(\right\{x\}-1)}$

where $\left\{x\right\}$ denotes the fractional part function is given by?

• A. $1$
• B. $\tan \left(1\right)$
• C. $\sin \left(1\right)$
• D. Doesn't exist

Answer 1

The correct option is D Doesn't exist

$\underset{x\to 0}{lim}\frac{\left(\tan \left(\right\{x\}-1)\right)\sin \left\{x\right\}}{\left\{x\right\}\left(\right\{x\}-1)}$
$RHL=\underset{x\to 0^{+}}{lim}\frac{\left(\tan \left(\right\{x\}-1)\right)\sin \left\{x\right\}}{\left\{x\right\}\left(\right\{x\}-1)}$
Since, $x\to 0^{+}\Rightarrow \left[x\right]\to 0$
$\Rightarrow \left\{x\right\}\to x$
So, $RHL=\underset{x\to 0^{+}}{lim}\frac{\left(\tan (x-1)\right)\sin x}{x(x-1)}$
$RHL=\tan 1$
Now, $LHL=\underset{x\to 0^{-}}{lim}\frac{\left(\tan \left(\right\{x\}-1)\right)\sin \left\{x\right\}}{\left\{x\right\}\left(\right\{x\}-1)}$
Since, $x\to 0^{-}\Rightarrow \left[x\right]\to -1$
$\Rightarrow \left\{x\right\}\to x+1$

$LHL=\underset{x\to 0^{-}}{lim}\frac{\tan x\sin (x+1)}{(x+1)\left(x\right)}$
$=\sin 1$
Since, $LHL\ne RHL$
Hence, limit does not exists.