Question

lim x tends to 0 sin2x(1-cos2x)/x^3

Answer 1

hello users ........

solution:-
we know that :
Lim x⇒0 [ (sin x ) / x ] = 1
and
Lim x ⇒0 [ (sin²x) / x² ] = 1 
and 
( 1 - cos 2x ) = 2 sin² x

Here,

lim x⇒0 [sin2x(1-cos2x)/x³ ] 

= Lim x⇒0 [(sin 2x) / x  × 2 (sin² x) / x² ] 

=  Lim x⇒0 [(sin 2x) / x]  × 2 Lim x⇒0 [ (sin² x) / x² ] 

=  Lim x⇒0 [ 2 (sin 2x) / 2 x]  × 2 Lim x⇒0 [ (sin² x) / x² ] 

= 2 Lim x⇒0 [(sin 2x) / 2x]  × 2 Lim x⇒0 [ (sin² x) / x² ] 

= 2×1 × 2 ×1= 4 

=> lim x⇒0 [sin2x(1-cos2x)/x³ ]  =  4 answer 

✪✪ hope it helps ✪✪

Answer 2

Answer:

10 thanxxx = 10 thanxx + follow ..............

15 thanxx = 20 thanxx + follow ..