\operatorname { lim } _ { x \rightarrow \pi / 6 } \frac { ( 2 \operato

1.	\operatorname { lim } _ { x \rightarrow \pi / 6 } \frac { ( 2 \operatorname { sin } ^ { 2 } x + \operatorname { sin } x - 1 ) } { ( 2 \operatorname { sin } ^ { 2 } x - 3 \operatorname { sin } x + 1 ) }
Answer» (2sin²x +sinx +1) = 2sin²x+2sinx-sinx-1= 2sinx(sinx+1)-1(sinx+1)= (2sinx-1)(sinx+1) 2sin²x-3sinx+1 = 2sin²x-2sinx-sinx+1=2sinx(sinx-1)-1(sinx-1)= (2sinx-1)(sinx-1) so, the fraction becomes limx→π/6 (2sinx-1)(sinx+1)/(2sinx-1)(sinx-1) =(sinx+1)/(sinx-1) now putting x = π/6 we get the limit as = (sin(π/6)+1)/(sin(π/6)-1) = (3/2)/(-1/2)= -3

\operatorname { lim } _ { x \rightarrow \pi / 6 } \frac { ( 2 \operatorname { sin } ^ { 2 } x + \operatorname { sin } x - 1 ) } { ( 2 \operatorname { sin } ^ { 2 } x - 3 \operatorname { sin } x + 1 ) }

Answer»

(2sin²x +sinx +1) = 2sin²x+2sinx-sinx-1= 2sinx(sinx+1)-1(sinx+1)= (2sinx-1)(sinx+1)

2sin²x-3sinx+1 = 2sin²x-2sinx-sinx+1=2sinx(sinx-1)-1(sinx-1)= (2sinx-1)(sinx-1)

so, the fraction becomes

limx→π/6 (2sinx-1)(sinx+1)/(2sinx-1)(sinx-1) =(sinx+1)/(sinx-1)

now putting x = π/6 we get the limit as

= (sin(π/6)+1)/(sin(π/6)-1) = (3/2)/(-1/2)= -3