Giải chi tiết giúp em vs câu b và c ạ

b) Ta xét

$\underset{x \to 0}{\lim} \dfrac{\cos(2x)-1}{\sin^2(3x)} = \underset{x \to 0}{\lim} \dfrac{-2\sin^2x}{\sin^2(3x)}$

$= -2\underset{x \to 0}{\lim}  \dfrac{\sin^2x}{x^2} . \dfrac{(3x)^2}{\sin^2(3x)} . \dfrac{1}{9}$

$= -\dfrac{2}{9} \underset{x \to 0}{\lim}  \dfrac{\sin^2x}{x^2} . \dfrac{(3x)^2}{\sin^2(3x)}$

Ta thấy rằng khi $x \to 0$ thì

$\underset{x \to 0}{\lim}  \dfrac{\sin^2x}{x^2}  = \underset{x \to 0}{\lim} \dfrac{(3x)^2}{\sin^2(3x)}=1$

Do đó giới hạn đã cho là

$\underset{x \to 0}{\lim} \dfrac{\cos(2x)-1}{\sin^2(3x)} = -\dfrac{2}{9}$.

c) Ta xét

$\underset{x \to 0}{\lim} \dfrac{\tan x - \sin x}{x^3} = \underset{x \to 0}{\lim} \dfrac{\sin x - \sin x \cos x}{x^3}$

$= \underset{x \to 0}{\lim} \dfrac{\sin x}{x} . \dfrac{1 - \cos x}{x^2}$

Ta lại có

$\underset{x \to 0}{\lim} \dfrac{\sin x}{x} = 1$. Do đó, giới hạn trở thành

$\underset{x \to 0}{\lim} \dfrac{\tan x - \sin x}{x^3} = \underset{x \to 0}{\lim} \dfrac{1 - \cos x}{x^2}$

$= \underset{x \to 0}{\lim} \dfrac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2}$

$= 2\underset{x \to 0}{\lim}  \dfrac{\sin^2 \left( \frac{x}{2} \right)}{\left( \frac{x}{2} \right)^2}. \dfrac{1}{4}$

Lại có

$\underset{x \to 0}{\lim}  \dfrac{\sin^2 \left( \frac{x}{2} \right)}{\left( \frac{x}{2} \right)^2} = 1$. Do đó

$\underset{x \to 0}{\lim} \dfrac{\tan x - \sin x}{x^3} = \dfrac{1}{2} $