Question

lim x=0. cos^2 /1-sinx​

Answer 1

Answer:

SOLUTION :-

GIVEN :-

$\sf \lim_{x \to \: 0 } \bigg( \frac{ \cos ^{2} (x) }{1 - \sin(x) } \bigg)$

$\sf \: Using \: \: the \: \: formula -$

$\blue{ \sf \: \cos^{2} (x) = 1 - \sin^{2} (x) }$

$\sf \lim_{x \to \: 0 } \bigg( \frac{1 - \sin^{2} (x) }{1 - \sin(x) } \bigg)$

$\sf \: Now \: use \: another \: formula -$

$\sf \blue{ \sf \: ( {x}^{2} - {y}^{2} ) = (x + y)(x - y)}$

$= \sf \lim_{x \to \: 0 } \bigg( \frac{ {1}^{2} - \sin^{2} (x) }{1 - \sin(x) } \bigg)$

$= \sf \lim_{x \to \: 0 } \bigg( \frac{ (1 - \sin(x) )(1 + \sin(x)) }{(1 - \sin(x)) } \bigg)$

$= \sf \lim_{x \to \: 0 } \bigg( \frac{ \cancel{ (1 - \sin(x) })(1 + \sin(x)) }{ \cancel{(1 - \sin(x)) }} \bigg)$

$= \sf \lim_{x \to \: 0 } \: (1 + \sin(x) )$

$\sf \: Now \: putting \: the \: limits$

$= \sf \: 1 + \sin(0)$

$= \sf \: 1 + 0$

$\sf \: = 1$