Question

Solve 12 and 13
No useless answer plz​

Answer 1

QUESTION 12:

$\displaystyle\lim_{x \to 0} \frac{ \sin x(1 - \cos x)}{ {x}^{3} }$

ANSWER 12:

$\displaystyle\lim_{x \to 0} \frac{ \sin x(1 - \cos x)}{ {x}^{3} } = \frac{1}{2}$

NEED TO USE L' HOSPITAL's RULE:

$\displaystyle\lim_{x \to 0} \frac{ \sin x(1 - \cos x)}{ {x}^{3} }$

$\dfrac{ \sin 0^{o} (1 - \cos 0^{o})}{ {0}^{3} }$

$\displaystyle\lim_{x \to 0} \frac{ \sin x(1 - \cos x)}{ {x}^{3} } = \frac{0}{0}=\infty$

HENCE WE NEED TO USE L' HOSPITAL'S RULE.

FORMULAE USED:

★ d / dx(sin x) = cos x

★ d / dx(cos x) = - sin x

★ d / dx(uv) = uv' + vu'

★ d / dx(x³) = 3x²

★ d / dx(x²) = 2x

★ (sin x) / x = (cos x) / x = (sin 2x) / 2x = 1

★ 2 sin x cos x = sin 2x

QUESTION 13:

$\displaystyle\lim_{x \to 0}\tan 5x \csc 4x$

ANSWER:

$\displaystyle\lim_{x \to 0}\tan 5x \csc 4x = \frac{5}{4}$

FORMULAE USED:

★ tan 5x / 5x = sin 4x / 4x = 1

★ cosec x = 1 / sin x

NOTE : REFER ATTACHMENT FOR EXPLANATION