Question

The value of $\underset{x\to 0}{lim}\frac{\left(\tan \right(\left\{x\right\}-1\left)\right)\sin \left\{x\right\}}{\left\{x\right\}\left(\right\{x\}-1)}$ where $\left\{x\right\}$ denotes the fractional part function

• A. $1$
• B. $\tan 1$
• C. $\sin 1$
• D. Is not existent

Answer 1

The correct option is D Is not existent

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

$LHL=\underset{h\to 0^{-}}{lim}\frac{\left(\tan \right((1-h)-1\left)\right)\sin (1-h)}{(1-h)\left(\right(1-h)-1)}$

$[\because \{x\}=\{0-h\}=1-h]$

$=\underset{h\to 0^{-}}{lim}\frac{-\tan h\sin (1-h)}{(1-h)(-h)}=\sin 1$

$RHL=\underset{h\to 0^{+}}{lim}\frac{\tan (h-1)\sin h}{h(h-1)}=\tan 1$

$\because LHL\ne RHL$

$\therefore$ limit does not exist