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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Sign up to join this communityEvaluate the limit without using L’Hospital’s rule and without using series expansion
$$\lim_{x \to 0} \frac{x - \sin x \cos x}{\tan x - x} $$
Rewrite $\sin x \cos x$ as $\frac{1}{2}\sin 2x$ and then use Taylor expansion for it and $\tan x$.
Since
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 $$\dfrac82\dfrac{\lim_{x\to0}\dfrac{2x-\sin2x}{(2x)^3}}{\lim_{x\to0}\dfrac{\tan x-x}{x^3}}$$
Now use Are all limits solvable without L'Hôpital Rule or Series Expansion