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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$\begingroup$ You've been here long enough that you should have anticipated this question, but ... What have you tried? $\endgroup$ – Blue Feb 16 '20 at 11:12
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Rewrite $\sin x \cos x$ as $\frac{1}{2}\sin 2x$ and then use Taylor expansion for it and $\tan x$.
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$\begingroup$ The Taylor expansion and Hospital they are just the same. $\endgroup$ – Michael Rozenberg Feb 16 '20 at 10:51
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$\begingroup$ I don’t agree. When using Taylor expansion, you don’t change the function inside limit, just assume it’s behavior near given point. And it has less requirements, then L’Hospital rule. $\endgroup$ – Yalikesifulei Feb 16 '20 at 11:52
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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 $$ \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x\cos x}} {{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{2} {3}x^3 + o(x^3 )}} {{\frac{1} {3}x^3 + o(x^3 )}} = 2 $$
answered Feb 16 '20 at 10:50
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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
answered Feb 16 '20 at 11:53
