Evaluating $\lim_{x \to 0} \frac{x - \sin x \cos x}{\tan x - x}$ without L'Hospital or series expansion

Evaluate the limit without using L’Hospital’s rule and without using series expansion

$$\lim_{x \to 0} \frac{x - \sin x \cos x}{\tan x - x}$$

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– Blue
Feb 16, 2020 at 11:12

Rewrite $$\sin x \cos x$$ as $$\frac{1}{2}\sin 2x$$ and then use Taylor expansion for it and $$\tan x$$.
1.$$\sin x\cos x = x - \frac{2} {3}x^3 + o(x^3 )$$
2.$$\tan x=x+\frac{x^3}{3}+o(x^3 )$$ You have that $$\mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x\cos x}} {{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{2} {3}x^3 + o(x^3 )}} {{\frac{1} {3}x^3 + o(x^3 )}} = 2$$
The limit can be expressed as $$\dfrac82\dfrac{\lim_{x\to0}\dfrac{2x-\sin2x}{(2x)^3}}{\lim_{x\to0}\dfrac{\tan x-x}{x^3}}$$