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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?

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    $\begingroup$ I'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. $\endgroup$ – user361424 Feb 28 '17 at 14:31
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    $\begingroup$ "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, $$x\sim\sin(\pi-x)$$ $\endgroup$ – Did Feb 28 '17 at 14:38
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    $\begingroup$ @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 $\tan x$). $\endgroup$ – user228113 Feb 28 '17 at 14:41
  • $\begingroup$ @G.Sassatelli, Did , Bernard , user361424 . Thankful $\endgroup$ – Almot1960 Feb 28 '17 at 14:43
  • $\begingroup$ @Almot1960, See math.stackexchange.com/questions/387333/… for $$\dfrac{\sin x-x}{x^3}$$ and $$\dfrac{\tan x-x}{x^3}$$ and divide them $\endgroup$ – lab bhattacharjee Mar 5 '17 at 13:28
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You can write the Sin and the Tan as exponential 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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  • $\begingroup$ Did you really need to expand at such an order? (I didn't downvote!) $\endgroup$ – Bernard Feb 28 '17 at 14:56
  • $\begingroup$ It 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. $\endgroup$ – kiara Feb 28 '17 at 14:58
  • $\begingroup$ But order $3$ suffices! In addition, as far as I know, there's no formulafor the general term in the power series expansion of $\tan x$. $\endgroup$ – Bernard Feb 28 '17 at 15:01
  • $\begingroup$ This is a perfectly good answer, why the downvote?! $\endgroup$ – Simply Beautiful Art Feb 28 '17 at 15:11

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