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Can one help finding this limit

$$\lim_{x \to 0}\frac{x^{3}-\sin^{2}x\tan x}{\tan(\sin x) - \sin (\tan x)}$$

L'Hospital's rule is permited.

(Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$)

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Idris Addou
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2 Answers2

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Hint. Recall that, for $x$ near $0$, you have $$\tan x = x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}+\mathcal{O}(x^9)$$ $$ \sin x = x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{17x^7}{720}+\mathcal{O}(x^9)$$ to obtain (using only relevant terms) $$\tan(\sin x) = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+\mathcal{O}(x^9)$$ $$\sin (\tan x) = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+\mathcal{O}(x^9)$$ $$x^{3}-\sin^{2}x\tan x = -\frac{x^7}{15}+\mathcal{O}(x^9)$$ Then the desired limit is equal to $-2$.

Euler's student
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Olivier Oloa
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  • Sorry - small correction. The coefficient of $x^7$ in $\tan(\sin x) - \sin(\tan x)$ is negative, so the limit is $-2$. – Andrey Kaipov Nov 17 '14 at 06:48
  • I am trying with l'Hospital's rule by adding and substractiong some things and to use some known limits as (sin(x)-x)/x^3 as x->0 for example. Thanks for your interest to Olivier and Andrey. – Idris Addou Nov 17 '14 at 07:36
  • Can one provide a solution by l'Hospital's Rule please? – Idris Addou Nov 18 '14 at 05:58
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Seven months ago, I have posted this question of computing the limit \begin{equation*} \lim_{x\rightarrow 0}\frac{x^{3}-\sin ^{2}\tan x}{\tan (\sin x)-\sin (\tan x) }, \end{equation*} without making use of Taylor series. After that, I have wrote the fraction as a product \begin{equation*} \frac{x^{3}-\sin ^{2}\tan x}{x^{7}}\times \frac{x^{7}}{\tan (\sin x)-\sin (\tan x)} \end{equation*} and I\ have asked the limit of the first fraction in another post $ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $

Paramanand has solved it and after I did it too in that post. However that technic of computations seems to do not apply for the second limit above. It follows that it remains to compute the second limit, or \begin{equation*} \lim_{x\rightarrow 0}\frac{\tan (\sin x)-\sin (\tan x)}{x^{7}}. \end{equation*} Recently, I have successfully computed a limit asked in the post A limit problem related to $\log \sec x$ where I\ developed some manipulations on limits. In this post, I am going to apply these manners of manipulations to compute this second limit and show that \begin{equation*} \lim_{x\rightarrow 0}\frac{\tan (\sin x)-\sin (\tan x)}{x^{7}}=\frac{1}{30}. \end{equation*} We shall use the following known limits \begin{equation*} \lim_{u\rightarrow 0}\frac{\tan u-u-\frac{1}{3}u^{3}-\frac{2}{15}u^{5}}{u^{7}% }=\frac{17}{315} \end{equation*} and \begin{equation*} \lim_{v\rightarrow 0}\frac{v-\frac{1}{6}v^{3}+\frac{1}{120}v^{5}-\sin v}{% v^{7}}=\frac{1}{5040}. \end{equation*} (Each one of this two limits can be computed with repeated use (7 times) of LHR).

The first step in the calculation process, is to convert the calculation in our limit, say $\lim_{x\rightarrow 0}\frac{f(x)}{x^{7}}$ into another limit say $\lim_{x\rightarrow 0}\frac{g(x)}{x^{7}}$ very simple comparatively to the original one. In the second step I will compute the second limit by just using limits which can be computed by LHR repeatedly, seven times to obtain the requested value. First, write \begin{eqnarray*} f(x) &=&\tan (\sin x)-\sin (\tan x) \\ &=&\left( \tan (\sin x)-\sin x-\frac{1}{3}\sin ^{3}x-\frac{2}{15}\sin ^{5}x\right) +\sin x+\frac{1}{3}\sin ^{3}x+\frac{2}{15}\sin ^{5}x \\ &&+\left( \tan x-\frac{1}{6}\tan ^{3}x+\frac{1}{120}\tan ^{5}x-\sin (\tan x)\right) -\tan x+\frac{1}{6}\tan ^{3}x-\frac{1}{120}\tan ^{5}x. \end{eqnarray*} Next, \begin{equation*} \left( \frac{\tan (\sin x)-\sin x-\frac{1}{3}\sin ^{3}x-\frac{2}{15}\sin ^{5}x}{x^{7}}\right) =\left( \frac{\tan (\sin x)-\sin x-\frac{1}{3}\sin ^{3}x-\frac{2}{15}\sin ^{5}x}{\sin ^{7}x}\right) \left( \frac{\sin x}{x}% \right) ^{7} \end{equation*} and since $\lim_{x\rightarrow 0}u(x)=\lim_{x\rightarrow 0}\sin x=0,$ then \begin{eqnarray*} \lim_{x\rightarrow 0}\left( \frac{\tan (\sin x)-\sin x-\frac{1}{3}\sin ^{3}x-% \frac{2}{15}\sin ^{5}x}{x^{7}}\right) &=&\lim_{x\rightarrow 0}\left( \frac{% \tan (\sin x)-\sin x-\frac{1}{3}\sin ^{3}x-\frac{2}{15}\sin ^{5}x}{\sin ^{7}x% }\right) \left( \frac{\sin x}{x}\right) ^{7} \\ &=&\left( \lim_{u\rightarrow 0}\frac{\tan u-u-\frac{1}{3}u^{3}-\frac{2}{15}% u^{5}}{u^{7}}\right) \left( \lim_{x\rightarrow 0}\frac{\sin x}{x}\right) ^{7} \\ &=&\left( \frac{17}{315}\right) \left( 1\right) ^{7}=\left( \frac{17}{315} \right) . \end{eqnarray*} The same way : \begin{equation*} \left( \frac{\tan x-\frac{1}{6}\tan ^{3}x+\frac{1}{120}\tan ^{5}x-\sin (\tan x)}{x^{7}}\right) =\left( \frac{\tan x-\frac{1}{6}\tan ^{3}x+\frac{1}{120} \tan ^{5}x-\sin (\tan x)}{\tan ^{7}x}\right) \left( \frac{\tan x}{x}\right) ^{7} \end{equation*}

and since $\lim_{x\rightarrow 0}v(x)=\lim_{x\rightarrow 0}\tan x=0,$ then \begin{eqnarray*} \lim_{x\rightarrow 0}\left( \frac{\tan x-\frac{1}{6}\tan ^{3}x+\frac{1}{120}% \tan ^{5}x-\sin (\tan x)}{x^{7}}\right) &=&\lim_{x\rightarrow 0}\left( \frac{\tan x-\frac{1}{6}\tan ^{3}x+\frac{1}{120}\tan ^{5}x-\sin (\tan x)}{% \tan ^{7}x}\right) \left( \frac{\tan x}{x}\right) ^{7} \\ &=&\left( \lim_{v\rightarrow 0}\frac{v-\frac{1}{6}v^{3}+\frac{1}{120}% v^{5}-\sin v}{v^{7}}\right) \left( \lim_{x\rightarrow 0}\frac{\tan x}{x} \right) ^{7} \\ &=&\left( \frac{1}{5040}\right) \left( 1\right) ^{7}=\frac{1}{5040}. \end{eqnarray*} It follows that

$$ \lim_{x\rightarrow 0}\frac{f(x)}{x^{7}}=\left( \frac{17}{315}\right) +\left(\frac{1}{5040}\right) +\lim_{x\rightarrow 0}\frac{g(x)}{x^{7}} \tag{k} $$

where $$ g(x)=+\sin x+\frac{1}{3}\sin ^{3}x+\frac{2}{15}\sin ^{5}x-\tan x+\frac{1}{6}% \tan ^{3}x-\frac{1}{120}\tan ^{5}x. $$

Now, taking into account the following limits which can be computed easily by repeated applications of LHR,

  1. $\lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}-\frac{1}{120}% x^{5}}{x^{7}}=-\frac{1}{5040}$
  2. $\lim_{x\rightarrow 0}\frac{\sin ^{3}x-x^{3}+\frac{1}{2}x^{5}}{x^{7}}=% \frac{13}{120}$
  3. $\lim_{x\rightarrow 0}\frac{\sin ^{5}x-x^{5}}{x^{7}}=-\frac{5}{6}$
  4. $\lim_{x\rightarrow 0}\frac{\tan x-x-\frac{1}{3}x^{3}-\frac{2}{15}x^{5}% }{x^{7}}=\frac{17}{315}$
  5. $\lim_{x\rightarrow 0}\frac{\tan ^{3}x-x^{3}-x^{5}}{x^{7}}=\frac{11}{15% }$
  6. $\lim_{x\rightarrow 0}\frac{\tan ^{5}x-x^{5}}{x^{7}}=\frac{5}{3}$.

we transform $g(x)$ as follows \begin{eqnarray*} g(x) &=&\left( \sin x-x+\frac{1}{6}x^{3}-\frac{1}{120}x^{5}\right) +\left( x-% \frac{1}{6}x^{3}+\frac{1}{120}x^{5}\right) \\ &&+\frac{1}{3}\left( \sin ^{3}x-x^{3}+\frac{1}{2}x^{5}\right) +\frac{1}{3}% \left( -x^{3}+\frac{1}{2}x^{5}\right) \\ &&+\frac{2}{15}\left( \sin ^{5}x-x^{5}\right) +\frac{2}{15}x^{5} \\ &&-\left( \tan x-x-\frac{1}{3}x^{3}-\frac{2}{15}x^{5}\right) +\left( -x-% \frac{1}{3}x^{3}-\frac{2}{15}x^{5}\right) \\ &&+\frac{1}{6}\left( \tan ^{3}x-x^{3}-x^{5}\right) +\frac{1}{6}\left( x^{3}+x^{5}\right) \\ &&-\frac{1}{120}\left( \tan ^{5}x-x^{5}\right) -\frac{1}{120}x^{5} \end{eqnarray*} and then write $\frac{g(x)}{x^{7}}$ as follows \begin{eqnarray*} \frac{g(x)}{x^{7}} &=&\left( \frac{\sin x-x+\frac{1}{6}x^{3}-\frac{1}{120}% x^{5}}{x^{7}}\right) +\left( \frac{x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}}{% x^{7}}\right) \\ &&+\frac{1}{3}\left( \frac{\sin ^{3}x-x^{3}+\frac{1}{2}x^{5}}{x^{7}}\right) -% \frac{1}{3}\left( \frac{-x^{3}+\frac{1}{2}x^{5}}{x^{7}}\right) \\ &&+\frac{2}{15}\left( \frac{\sin ^{5}x-x^{5}}{x^{7}}\right) +\frac{2}{15}% \frac{x^{5}}{x^{7}} \\ &&-\left( \frac{\tan x-x-\frac{1}{3}x^{3}-\frac{2}{15}x^{5}}{x^{7}}\right) +\left( \frac{-x-\frac{1}{3}x^{3}-\frac{2}{15}x^{5}}{x^{7}}\right) \\ &&+\frac{1}{6}\left( \frac{\tan ^{3}x-x^{3}-x^{5}}{x^{7}}\right) +\frac{1}{6} \left( \frac{x^{3}+x^{5}}{x^{7}}\right) \\ &&-\frac{1}{120}\left( \frac{\tan ^{5}x-x^{5}}{x^{7}}\right) -\frac{1}{120} \frac{x^{5}}{x^{7}} \end{eqnarray*} Remark that \begin{equation*} \left( \tfrac{x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}}{x^{7}}\right) -\frac{1}{% 3}\left( \tfrac{-x^{3}+\frac{1}{2}x^{5}}{x^{7}}\right) +\frac{2}{15}\tfrac{% x^{5}}{x^{7}}+\left( \tfrac{-x-\frac{1}{3}x^{3}-\frac{2}{15}x^{5}}{x^{7}}% \right) +\frac{1}{6}\left( \tfrac{x^{3}+x^{5}}{x^{7}}\right) -\frac{1}{120}% \tfrac{x^{5}}{x^{7}}=0 \end{equation*} so \begin{eqnarray*} \frac{g(x)}{x^{7}} &=&\left( \frac{\sin x-x+\frac{1}{6}x^{3}-\frac{1}{120}% x^{5}}{x^{7}}\right) +\frac{1}{3}\left( \frac{\sin ^{3}x-x^{3}+\frac{1}{2}% x^{5}}{x^{7}}\right) \\ &&+\frac{2}{15}\left( \frac{\sin ^{5}x-x^{5}}{x^{7}}\right) -\left( \frac{% \tan x-x-\frac{1}{3}x^{3}-\frac{2}{15}x^{5}}{x^{7}}\right) \\ &&+\frac{1}{6}\left( \frac{\tan ^{3}x-x^{3}-x^{5}}{x^{7}}\right) -\frac{1}{% 120}\left( \frac{\tan ^{5}x-x^{5}}{x^{7}}\right) \end{eqnarray*} and by passing to the limit one obtains \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{g(x)}{x^{7}} &=&\left( -\frac{1}{5040}\right) +% \frac{1}{3}\left( \frac{13}{120}\right) +\frac{2}{15}\left( -\frac{5}{6}% \right) -\left( \frac{17}{315}\right) +\frac{1}{6}\left( \frac{11}{15}% \right) -\frac{1}{120}\left( \frac{5}{3}\right) \\ &=&-\frac{1}{48}. \end{eqnarray*}

Using (k) it follows that $$ \lim_{x\rightarrow 0}\frac{f(x)}{x^{7}}=\left( \frac{17}{315}\right) +\left( \frac{1}{5040}\right) +\left( -\frac{1}{48}\right) =\frac{1}{30}. $$

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Idris Addou
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  • On the surface this does look like LHR, but is actually the Taylor series approach only. The smartness used here is in avoiding calculating Taylor series for $\tan(\sin x)$ and instead deal with $\tan s$ with $s = \sin x$ and replace denominator $x$ by $\sin x = s$. – Paramanand Singh Jun 18 '15 at 10:36
  • The Taylor's theorem which is normally used for limit calculation says that if $f^{(n)}(a)$ exists then $$f(a + h) = f(a) + hf'(a) + \cdots + \frac{f^{(n)}(a)}{n!}h^{n} + o(h^{n})$$ and its proof is given by establishing the limit formula $$\lim_{h \to 0}\dfrac{f(a + h) - f(a) - hf'(a) - \cdots - \dfrac{f^{(n - 1)}(a)}{(n - 1)!}h^{n - 1}}{h^{n}} = \frac{f^{(n)}(a)}{n!}$$ and this limit is calculated via repeated application of LHR. See more details at http://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-4.html – Paramanand Singh Jun 18 '15 at 10:36
  • ok and so, Is it good or not? – Idris Addou Jun 18 '15 at 10:41
  • I would say it is good, but perhaps using Taylor would have taken less space and time. I suppose the limit of $g(x)/x^{7}$ could be calculated by some other manipulation instead of using Taylor expansions. I will try to think about $g(x)$ a bit more. Right now I can only say that the standard limit 3 and 6 for $g(x)$ can be easily caculated by factoring. Like $$\frac{\sin^{5}x - x^{5}}{x^{7}} = \frac{\sin x - x}{x^{3}}\cdot\left(\frac{\sin^{4}x}{x^{4}} + \frac{\sin^{3}x}{x^{3}} + \frac{\sin^{2}x}{x^{2}} + \frac{\sin x}{x} + 1\right) \to -\frac{1}{6}\cdot 5$$ – Paramanand Singh Jun 18 '15 at 11:22
  • I also tried my hand at this problem (finally!) See http://math.stackexchange.com/a/1332331/72031 – Paramanand Singh Jun 20 '15 at 07:25