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Can someone show me how is possible to prove that \begin{equation*} \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} \end{equation*} but without Taylor series. One can use L'Hospital rule if necessary. I was not able.

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    $\begingroup$ Strictly speaking you don't need Taylor series, just the few terms, which you can compute by differentiating $\sin$ and $\tan$ a few times each. This is certainly more appealing than differentiating the expression in the limit seven times... $\endgroup$ – Travis May 19 '15 at 5:43
  • $\begingroup$ L'Hospital rule would be a real nightmare to me. As Travis said, very few terms are required in the series. $\endgroup$ – Claude Leibovici May 19 '15 at 5:50
  • $\begingroup$ Maybe I have not asked my question correctly. It is by using just few terms that one can obtain easily the limit $\frac{1}{15}$, but the challenge is in to avoid using any expansion. $\endgroup$ – Idris May 19 '15 at 6:01
  • $\begingroup$ There is one thing i have learnt while solving limit problems on MSE is that if you really need to use LHR or Taylor (meaning it is not possible to do the limit via algebra of limits and sqeeze theorem) then it is better to make some algebraical manipulations before applying LHR or Taylor. In case of LHR we need to make algebraic manipulation even after its application to avoid repeated use of LHR. See my answer here which just manages to do the problem by applying LHR 3 times. It saves the labor of dealing with complicated expressions. $\endgroup$ – Paramanand Singh May 19 '15 at 7:49
  • $\begingroup$ @Paramanand. Thank you very much. Your method is powerfull. After reading it, I\ am able now to combine it with some unpublished computations I\ did. My favorite method is to do not use LHR nor Taylor expansion whenever possible and to came back all the computations to the basic limits only. $\endgroup$ – Idris May 19 '15 at 20:41
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Let the desired limit be denoted by $L$.

We have via LHR $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\tag{1}$$ and $$\lim_{x \to 0}\frac{\tan x - x}{x^{3}} = \lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}} = \frac{1}{3}\tag{2}$$ and we also have $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{3}$$ Multiplying the 3 limits above we get \begin{align} &\lim_{x \to 0}\frac{(x - \sin x)(\tan x - x)\sin x}{x^{7}} = \frac{1}{18}\notag\\ &\Rightarrow\lim_{x \to 0}\frac{(x\sin x\tan x - x^{2}\sin x - \sin^{2}x\tan x + x\sin^{2}x)}{x^{7}} = \frac{1}{18}\notag\\ &\Rightarrow\lim_{x \to 0}\frac{x\sin x\tan x - x^{2}\sin x - x^{3} + x^{3} - \sin^{2}x\tan x + x\sin^{2}x}{x^{7}} = \frac{1}{18}\notag\\ &\Rightarrow\lim_{x \to 0}\frac{x\sin x\tan x - x^{2}\sin x + x\sin^{2}x - x^{3}}{x^{7}} - \frac{\sin^{2}x\tan x - x^{3}}{x^{7}} = \frac{1}{18}\notag\\ &\Rightarrow\lim_{x \to 0}\frac{\sin x\tan x - x\sin x + \sin^{2}x - x^{2}}{x^{6}} - L = \frac{1}{18}\notag\\ \end{align} Our job is done if we can show that $$\lim_{x \to 0}\frac{\sin x\tan x - x\sin x + \sin^{2}x - x^{2}}{x^{6}} = \frac{11}{90}\tag{4}$$ Multiplying $(1)$ and $(2)$ we get $$\lim_{x \to 0}\frac{x\tan x - x^{2} - \sin x\tan x + x\sin x}{x^{6}}= \frac{1}{18}\tag{5}$$ Adding $(4)$ and $(5)$ we see that our proof is complete if we show that $$\lim_{x \to 0}\frac{x\tan x + \sin^{2}x - 2x^{2}}{x^{6}} = \frac{8}{45}\tag{6}$$ Squaring $(1)$ we get $$\lim_{x \to 0}\frac{\sin^{2}x + x^{2} - 2x\sin x}{x^{6}} = \frac{1}{36}\tag{7}$$ Subtracting $(7)$ from $(6)$ we see that proof is complete if we show that $$\lim_{x \to 0}\frac{\tan x + 2\sin x - 3x}{x^{5}}= \frac{3}{20}\tag{8}$$ It is this limit which we will calculate using LHR as follows \begin{align} A &= \lim_{x \to 0}\frac{\tan x + 2\sin x - 3x}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\sec^{2} x + 2\cos x - 3}{5x^{4}}\text{ (apply LHR)}\notag\\ &= \lim_{x \to 0}\frac{1 + 2\cos^{3} x - 3\cos^{2}x}{5x^{4}\cos^{2}x}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\frac{1 + 2\cos^{3} x - 3\cos^{2}x}{x^{4}}\notag\\ &= \frac{1}{5}\lim_{x \to 0}\frac{(\cos x - 1)^{2}(2\cos x + 1)}{x^{4}}\notag\\ &= \frac{3}{5}\lim_{x \to 0}\left(\frac{1 - \cos x}{x^{2}}\right)^{2}\notag\\ &= \frac{3}{5}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{3}{20} \end{align}

Thus the proof is complete by application of LHR three times (once in proof of $(1)$, $(2)$, $(8)$ each). Also note that if you know the result $(1)$ then result $(2)$ can be derived from $(1)$ by subtraction and noting that the limit $$\lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}$$ can be calculated without LHR very easily. See this question. So in reality we only need two application of LHR for this problem.

Update: While dealing with limit expressions of type $\lim_{x \to 0}f(x)/x^{n}$ for large $n$ (here $n = 7$), I have often found it useful to multiply several well knows limits of type $g(x)/x^{m}$ with smaller values of $m$ to get something like $h(x)/x^{n}$. Expectation is that some terms of $f(x)$ match with those of $h(x)$ and a subtraction would cancel these terms. Also it is expected that resulting expression will be simplified to $p(x)/x^{r}$ when $r < n$. Continue this till we get very small values of exponent of $x$ in denominator. Here for example I have reduced an expression with $x^{7}$ to finally an expression with $x^{5}$ in denominator. See this technique applied to $$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$$ here. Another application of the same technique can be found here as well.

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  • $\begingroup$ Why the downvote? $\endgroup$ – Paramanand Singh May 7 '16 at 2:48
  • $\begingroup$ I do not think i did a downvote, in the contrary i appreciate your solution and i learned from it $\endgroup$ – Idris May 7 '16 at 2:56
  • $\begingroup$ I know Idris. My question is directed to the anonymous downvoter who did not leave any comment regarding his downvote. $\endgroup$ – Paramanand Singh May 7 '16 at 2:58
  • $\begingroup$ Ok. By the way i did some vidéo on YouTube where i explain this usage of the basic limits: it is entitled ´Règle de l'hospital: usage astucieux' it is in french $\endgroup$ – Idris May 7 '16 at 3:13
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@Paramanand. Thank you very much. Your method is powerfull. After reading it, I am able now to combine it with some unpublished computations I did. My favorite method is to do not use LHR nor Taylor expansion whenever possible and to came back all the computations to the basic limits only. For example, to compute the limit \begin{equation*} \lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^{3}} \end{equation*} one can make use the following basic limits \begin{equation*} \lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}=\frac{1}{3},\ \ \ \ \ \ \text{and }\ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6}, \end{equation*} as follows \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^{3}} &=&\lim_{x\rightarrow 0}% \frac{(\tan x-x)+(x-\sin x)}{x^{3}} \\ &=&\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}+\lim_{x\rightarrow 0}\frac{% x-\sin x}{x^{3}} \\ &=&\frac{1}{3}+\frac{1}{6}=\frac{1}{2}. \end{eqnarray*} To compute the limit $A=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\tan x+2\sin x-3x}{x^{5}}$ I suggest the same way but to push further one more order. By LHR one can use the following basic limits: \begin{equation*} \lim_{x\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})}{x^{5}}=\frac{2}{% 15},\ \ \ \ \ \ and\ \ \ \ \ \ \ \lim_{x\rightarrow 0^{+}}\frac{\sin x-x+% \frac{1}{6}x^{3}}{x^{5}}=\frac{1}{120}. \end{equation*}

Hence,

\begin{eqnarray*} \lim_{x\rightarrow 0^{+}}\frac{\tan x+2\sin x-3x}{x^{5}} &=&\lim_{x% \rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})+2(\sin x-x+\frac{1}{6}% x^{3})}{x^{5}} \\ &=&\lim_{x\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})}{x^{5}}% +2\lim_{x\rightarrow 0^{+}}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} \\ &=&\frac{2}{15}+2\left( \frac{1}{120}\right) \\ &=&\frac{3}{20}. \end{eqnarray*}

Now I can say that taking into account the development of Paramanand Singh, and the remark above, it is (finally) possible to compute the original limit by making use of only the basic following limits \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\ \lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3} \\ \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\ \lim_{x\rightarrow 0}\frac{\tan x-x-\frac{1}{3}x^{3}}{x^{5}} &=&\frac{2}{15} \\ \lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} &=&\frac{1}{120}% . \end{eqnarray*} The computations could be conducted as follows \begin{eqnarray*} \frac{\sin ^{2}x\tan x-x^{3}}{x^{7}} &=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right) \left( \frac{\sin x}{x}\right) +\left( \frac{\sin x\tan x-x\sin x+\sin ^{2}x-x^{2}}{x^{6}}\right) \\ &=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right) \left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right) \\ &&+\left( \frac{x\tan x+\sin ^{2}x-2x^{2}}{x^{6}}\right) \\ &=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right) \left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right) \\ &&+\left( \frac{\sin x-x}{x^{3}}\right) ^{2}+\left( \frac{\tan x+2\sin x-3x}{% x^{5}}\right) \\ &=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right) \left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right) \\ &&+\left( \frac{\sin x-x}{x^{3}}\right) ^{2}+\left( \frac{\tan x-x-\frac{1}{3% }x^{3}}{x^{5}}\right) +2\left( \frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}}% \right) \end{eqnarray*} Therefore, \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}} &=&\left( \frac{-1% }{6}\right) \left( \frac{1}{3}\right) \left( \frac{1}{1}\right) +\left( \frac{-1}{6}\right) \left( \frac{1}{3}\right) +\left( \frac{-1}{6}\right) \left( \frac{-1}{6}\right) +\left( \frac{2}{15}\right) +2\left( \frac{1}{120}% \right) \\ &=&\frac{1}{15}. \end{eqnarray*}

It is like decomposing a non-prime number into a product of prime numbers...Here, the basic limits are the prime numbers, and performing 'a' decomposition is a state of art. Thanks again to Paramanand.

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  • $\begingroup$ Good to see that you are using this technique quite nicely. But remember to minimize the use of LHR. For example the list of basic limits which you have mentioned would require 3 times LHR in total. And the computation which you have done to express $(\sin^{2}x\tan x - x^{3})/x^{7}$ in terms of other limits are not so obvious. It is better to start from original expression and try to show its dependency on limit of simpler expressions by making use of the already available limits. Note however that if you get such problems in an exam, best option is to use Taylor series to save time. $\endgroup$ – Paramanand Singh May 20 '15 at 5:39
  • $\begingroup$ It is not for an exam that i do this computations! I am not a student! I have students! $\endgroup$ – Idris May 20 '15 at 5:43
  • $\begingroup$ If i can send you an email it would be better. $\endgroup$ – Idris May 20 '15 at 5:44
  • $\begingroup$ Sorry I did not get much info from your profile. So bit confused. Since you have students it would be great if you can teach this technique to them. BTW its bit difficult to give here my email publicly. Please be available in math chat room. there i can discuss more. $\endgroup$ – Paramanand Singh May 20 '15 at 5:49
  • $\begingroup$ chat.stackexchange.com/rooms/36/mathematics $\endgroup$ – Idris May 20 '15 at 5:52
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L'Hospital once the fraction becomes

$$\frac{2\sin^2{x}+\tan^2{x}-3x^2}{7x^6}$$

Still an indeterminate

L'hospital 6 more times and rearranging the fraction becomes

$$\frac{4\cos{2x}}{315}+\sec^8{x}-\frac{4\sec^6{x}}{3}+\frac{2\sec^4{x}}{5}-\frac{4\sec^2{x}}{315}$$

And the limit for $x\to 0$ is $1-\frac{4}{3}+\frac{2}{5}=\frac{1}{15}$

7 times l'Hospital is so much of a pain: Taylor in this case is more convenient

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