limh →0 (2[√3 sin((π/6)+h)- cos((π/6)+h)]/√3h(√3 cos h-sin h)) equal to

Question

limh →0 (2[√3 sin((π/6)+h)- cos((π/6)+h)]/√3h(√3 cos h-sin h)) equal to (A) (2/3) (B) (4/3) (C) -2√3 (D) -(4/3)

Tardigrade Admin · Accepted Answer

limh→0 (2[√3 sin((π/6)+h) - cos((π/6)+h)]/√3 h (√3 cos h - sin h)) = lim h→ 0 ([√3 ((1/2) cos h +(√3/2) sin h) -((√3/2) cos h - (1/2) sin h )]/√3 h (√3 cos h - sin h)) = lim h→ 0 (√3 cos h +3 sin h - √3 cos h + sin h /√3 h[√3 cos h - sin h]) = lim h→ 0 ( 4 ( sin h/h)/√3 [√3 cos h - sin h] ) = (4/√3 (√3 -0) ) = (4/3)

Q. $lim_{h \to 0} \frac{2 [ 3 s in ( \frac{π}{6} + h ) - cos ( \frac{π}{6} + h ) ]}{3 h ( 3 cos h - s in h )}$ equal to

$\frac{2}{3}$

$\frac{4}{3}$

$- 23$

$- \frac{4}{3}$

Q. limh→0​3​h(3​cosh−sinh)2[3​sin(6π​+h)−cos(6π​+h)]​ equal to

32​

34​

−23​

−34​

Q. $lim_{h \to 0} \frac{2 [ 3 s in ( \frac{π}{6} + h ) - cos ( \frac{π}{6} + h ) ]}{3 h ( 3 cos h - s in h )}$ equal to

$\frac{2}{3}$

$\frac{4}{3}$

$- 23$

$- \frac{4}{3}$