Find the limits without L'Hôpital's rule $$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $$ My Try: $$\lim_{ x \to 0 }\frac{\sin(\pi-x)-\sin x}{\tan(\pi+x)-\tan x}=?\\\lim_{ x \to 0 }\frac{2\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2})}{\frac{\sin(\frac{π}{2}-x)}{\cos(\pi+x)\cos(x)}}=\lim_{ x \to 0 }\frac{(2\cos x)(-\cos x)(\cos(\frac{\pi}{2}))}{\cos x}=0$$ but: $$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=-1/2$$ Where is my mistake?
Find the limits without L'Hôpital:$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $
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1I'm not sure what you think you're doing at the first step, but my guess would be your error is that you've split the limit of an indeterminate form, which you generally can't do. – 2017-02-28
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1"Where is my mistake?" At your very first step, when you replace $$x-\sin x$$ by $$\sin(\pi-x)-\sin x$$ Note that the former is of order $x^3$ while the latter is identically $0$, hence replacing the former by the latter is a sure way to obtain absurd consequences... although, yes, $$x\sim\sin(\pi-x)$$ – 2017-02-28
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1@almot1960 You might have missed it, but substituting $x$ with $\sin(\pi-x)$ is the same as substituting $x$ with $\sin x$. Not only in the sense that it is the same mistake, but also in the sense that it's the same *function*, since $\sin x=\sin(\pi-x)$. Same goes for $\tan(\pi+x)$ (also known as $\tan x$). – 2017-02-28
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0@G.Sassatelli, Did , Bernard , user361424 . Thankful – 2017-02-28
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0@Almot1960, See http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion for $$\dfrac{\sin x-x}{x^3}$$ and $$\dfrac{\tan x-x}{x^3}$$ and divide them – 2017-03-05
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0See [here](http://rgmia.org/papers/v8n3/Huygens.pdf).Furthermore there is a proof with a geometric approach but I didn't find it :). – 2017-09-26
1 Answers
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You can write the Sin and the Tan as expoential series, and then you get
$$x-\sin(x)=\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}-\frac{x^9}{362880}+O\left(x^{11}\right )$$
and
$$-\frac12 (x-\tan(x))=\frac{x^3}{6}+\frac{x^5}{15}+\frac{17 x^7}{630}+\frac{31 x^9}{2835}+O\left(x^{11}\right)$$
If you divide the two sums you get $$ 1-\frac{9 x^2}{20}+\frac{27 x^4}{1400}-\frac{27 x^6}{56000}+\frac{201 x^8}{43120000}+O\left(x^{10}\right)$$ which goes clearly to 1.
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0Did you really need to expand at such an order? (I didn't downvote!) – 2017-02-28
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0It would be better to not use this notation at all but sum to infinity. As $${\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}\\[8pt]\end{aligned}}$$ And equivalently with tan. – 2017-02-28
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0But order $3$ suffices! In addition, as far as I know, there's no formulafor the general term in the power series expansion of $\tan x$. – 2017-02-28
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0This is a perfectly good answer, why the downvote?! – 2017-02-28