displaystyle limx → 0((1 - cos 2x)(3 + cos x)/x tan 4x) is equal to

Question

displaystyle limx → 0((1 - cos 2x)(3 + cos x)/x tan 4x) is equal to (A) 1/2 (B) 1 (C) 2 (D) -1/4

Tardigrade Admin · Accepted Answer

limx→0((1 - cos 2x)(3 + cos x)/x( tan 4x)) = lim x→ 0(1 - cos 2x/x2(( tan 4x)x)) ( 3 + cos x) = lim x→ 0((2 sin2x/x2)). ((x/ tan 4x)) (3 + cos x ) = 2 ×(1/4) ×4 = 2

Q. $x \to 0 lim \frac{( 1 - cos 2 x ) ( 3 + cos x )}{x tan 4 x}$ is equal to

1/2

1

2

-1/4

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x ( t a n 4 x )}$
$= lim_{x \to 0} \frac{1 - c o s 2 x}{x ^{2} ( \frac{t a n 4 x}{x} )} (3 + cos x)$
$= lim_{x \to 0} (\frac{2 s i n ^{2} x}{x ^{2}}) . (\frac{x}{t a n 4 x}) (3 + cos x)$
$= 2 \times \frac{1}{4} \times 4 = 2$

Q. x→0lim​xtan4x(1−cos2x)(3+cosx)​ is equal to

1/2

1

2

-1/4

limx→0​x(tan4x)(1−cos2x)(3+cosx)​ =limx→0​x2(xtan4x​)1−cos2x​(3+cosx) =limx→0​(x22sin2x​).(tan4xx​)(3+cosx) =2×41​×4=2

Q. $x \to 0 lim \frac{( 1 - cos 2 x ) ( 3 + cos x )}{x tan 4 x}$ is equal to

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x ( t a n 4 x )}$
$= lim_{x \to 0} \frac{1 - c o s 2 x}{x ^{2} ( \frac{t a n 4 x}{x} )} (3 + cos x)$
$= lim_{x \to 0} (\frac{2 s i n ^{2} x}{x ^{2}}) . (\frac{x}{t a n 4 x}) (3 + cos x)$
$= 2 \times \frac{1}{4} \times 4 = 2$