Question

limit x > 0 (tanx-sin3x)/x^3

Answer 1

Step-by-step explanation:

We have,

$\lim_{x \rarr0} \frac{ \tan(x) - \sin(3x) }{ {x}^{3} }$

Since, it is 0/0 form.. using l'hospital rule,

$= \lim_{x \rarr0} \frac{ \sec ^{2} (x) - 3\cos(3x) }{ 3{x}^{2} }$

$= \lim_{x \rarr0} \frac{ 2\sec ^{2} (x) \tan(x) + 9\sin(3x) }{ 6x }$

$= \lim_{x \rarr0} \frac{ 2\sec ^{2} (x) \tan(x) }{ 6x } + \lim _{x \rarr0} \frac{9 \sin(3x) }{6x}$

$= \lim_{x \rarr0} \sec^{2} (x). \lim_{x \rarr0} \frac{ \tan(x) }{ 3x } + \frac{9}{2} \lim _{3x \rarr0} \frac{ \sin(3x) }{3x}$

$= 1 \times \frac{1}{3} + \frac{9}{2}$

$= \frac{1}{3} + \frac{9}{2}$

$= \frac{2 + 27}{6}$

$= \frac{29}{6}$