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  4. limx → 0 (sin [cosx] /1+[cosx]) =

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Q. $\lim_{x \to 0} \frac {sin [cosx] } {1+[cosx]} $=

Solution:

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