Question

${\lim }_{x\to 0}\frac{sin[cosx]}{1+[cosx]}$=

• A. 1
• B. 0
• C. does not exist
• D. none of these

Answer 1

Answer: (B)

${\lim }_{x\to 0}\frac{sin[cosx]}{1+[cosx]}=0$
[$\because$ cos $x$ lies between 0 and lin the IVth quadrant
$\therefore$ [cos $x$] = 0 $\Rightarrow$ sin [cos $x$] = 0]
${\lim }_{x\to 0+}\frac{sin[cosx]}{1+[cosx]}=0$
[$\because$ cos $x$ lies between 0 and 1 in the first quadrant]
${\lim }_{x\to 0}\frac{sin[cosx]}{1+[cosx]}=0$