Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $$f(x) = \dfrac{{\displaystyle 3\int_{0}^{x}(1 + \sec t)\log\sec t\,dt}}{(\log\sec x)\{x + \log(\sec x + \tan x)\}}$$ then prove that $$\lim_{x \to {\pi/2}^{-}}f(x) = \frac{3}{2}$$ and $$\lim_{x \to 0}\frac{f(x) - 1}{x^{4}} = \frac{1}{420}$$

Looking at the integral sign in numerator I see that the best way to attack this problem is via L'Hospital Rule. But that requires to show that the integral diverges to $\infty$ as $x \to {\pi/2}^{-}$. Assuming that this is the case I solved the first limit by applying L'Hospital's rule twice. But for the second limit it seems hopeless to try L'Hospital because of denominator $x^{4}$ which might require 4 times its application.

Looking at the functions involved it does not look easy to apply Taylor's series expansions. I am not sure if there is any elegant solution for the second problem. Please let me know any hints or a solution to the second limit.

Update: I tried some simplification along with LHR for the second limit but still the final solution is eluding.

Let $a(x), b(x)$ be the numerator and denominator of $f(x)$. Clearly we can see that \begin{align} B &= \lim_{x \to 0}\frac{b(x)}{x^{3}}\notag\\ &= \lim_{x \to 0}\frac{\log\sec x\{x + \log(\sec x + \tan x)\}}{x^{3}}\notag\\ &= -\lim_{x \to 0}\frac{\log\cos x\{x + \log(1 + \sin x) - \log \cos x\}}{x^{3}}\notag\\ &= -\lim_{x \to 0}\frac{\log\cos x}{x^{2}}\cdot\frac{x + \log(1 + \sin x) - \log \cos x}{x}\notag\\ &= -\lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{\cos x - 1}\cdot\frac{\cos x - 1}{x^{2}}\cdot\frac{x + \log(1 + \sin x) - \log \cos x}{x}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\frac{x + \log(1 + \sin x) - \log \cos x}{x}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\left(1 + \frac{\log(1 + \sin x)}{\sin x}\cdot\frac{\sin x}{x} - \frac{\log (1 + \cos x - 1)}{\cos x - 1}\cdot x\cdot \frac{\cos x - 1}{x^{2}}\right)\notag\\ &= \frac{1}{2}\cdot 2 = 1\notag \end{align} Thus we can write \begin{align} L &= \lim_{x \to 0}\frac{f(x) - 1}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{a(x) - b(x)}{b(x)x^{4}}\notag\\ &= \lim_{x \to 0}\frac{a(x) - b(x)}{x^{7}}\cdot\frac{x^{3}}{b(x)}\notag\\ &= \lim_{x \to 0}\frac{a(x) - b(x)}{x^{7}}\notag\\ &= \frac{1}{7}\lim_{x \to 0}\frac{a'(x) - b'(x)}{x^{6}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{7}\lim_{x \to 0}\frac{3(1 + \sec x)\log\sec x - \tan x\{x + \log(\sec x + \tan x)\} -\log\sec x\{1 + \sec x\}}{x^{6}}\notag\\ &= \frac{1}{7}\lim_{x \to 0}\frac{2(1 + \sec x)\log\sec x - \tan x\{x + \log(\sec x + \tan x)\}}{x^{6}}\notag\\ &= \frac{1}{7}\lim_{x \to 0}\frac{2(1 + \cos x)\log\sec x - \sin x\{x + \log(\sec x + \tan x)\}}{x^{6}\cos x}\notag\\ &= \frac{1}{7}\lim_{x \to 0}\frac{2(1 + \cos x)\log\sec x - \sin x\{x + \log(\sec x + \tan x)\}}{x^{6}}\notag\\ \end{align} I wonder what could be done to go further.

share|cite|improve this question
Would the downvoter care to comment? Let me know if I am missing something or if there is any room for improvement in the question. – Paramanand Singh Jun 15 at 16:45
I don't know why there was a down vote. However, it is not clear what the brackets mean in the denominator. Do you mean the denominator to be $\left(\log\sec x\right)\left(x + \log(\sec x + \tan x)\right)$? Or $\log\sec \left(x(x + \log(x\sec x + \tan x))\right)$? – Michael Jun 17 at 5:12
@Michael: it means $(\log \sec x)(x + \log(\sec x + \tan x))$ I will update the question to reflect that – Paramanand Singh Jun 17 at 5:15
@Paramanand. It seems to me that the last two lines should be as follows: \begin{eqnarray*} &=&\frac{1}{7}\lim_{x\rightarrow 0}\frac{2\left( \cos x\right) (1+\cos x)\log \sec x-\sin x\left( x+\log (\sec x+\tan x)\right) }{x^{6}\cos x} \\ &=&\frac{1}{7}\lim_{x\rightarrow 0}\frac{2\left( \cos x\right) (1+\cos x)\log \sec x-\sin x\left( x+\log (\sec x+\tan x)\right) }{x^{6}}. \end{eqnarray*} Can you confirm please? – Idris Jun 17 at 5:39
@Idris: If you see 3rd last line it begins with $2(1 + \sec x)$ and by multiplying with $\cos x$ it becomes $2 (1 + \cos x)$ so my calculation seems ok. – Paramanand Singh Jun 17 at 5:45

3 Answers 3

up vote 3 down vote accepted

Let \begin{eqnarray*} g(x) &=&2(1+\cos x)\log \sec x-\sin x\{x+\log (\sec x+\tan x)\} \\ &=&\{\sin x-2(1+\cos x)\}\log \cos x-x\sin x-\sin x\log (1+\sin x) \\ &=&\{\sin x-2(1+\cos x)\}A(x)-B(x)-C(x) \end{eqnarray*} where \begin{eqnarray*} A(x) &=&\log \cos x \\ B(x) &=&x\sin x \\ C(x) &=&\sin x\log (1+\sin x) \end{eqnarray*}

Let us start with $B(x).$ \begin{eqnarray*} B(x) &=&x\sin x \\ &=&x\left[ \sin x-x+\frac{1}{6}x^{3}+x-\frac{1}{6}x^{3}\right] \\ &=&x\left[ \sin x-x+\frac{1}{6}x^{3}\right] +x^{2}-\frac{1}{6}x^{4} \\ &=&x^{6}\left( \frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}}\right) +x^{2}-\frac{1% }{6}x^{4}. \end{eqnarray*} Now let us consider in the same lines $A(x)$ \begin{eqnarray*} A(x) &=&\log \cos x \\ &=&\log (1+(\cos x-1))-(\cos x-1)+\frac{1}{2}(\cos x-1)^{2} \\ &&+(\cos x-1)-\frac{1}{2}(\cos x-1)^{2} \\ &=&\left( \lim_{x\rightarrow 0}\frac{\log (1+(\cos x-1))-(\cos x-1)+\frac{1}{% 2}(\cos x-1)^{2}}{(\cos x-1)^{3}}\right) \left( \frac{\cos x-1}{x^{2}}% \right) ^{3}\left( x^{6}\right) \\ &&+(\cos x-1)-\frac{1}{2}(\cos x-1)^{2} \end{eqnarray*} It remains $C(x)$ \begin{eqnarray*} C(x) &=&\sin x\log (1+\sin x) \\ &=&\sin x(\log (1+\sin x)-\sin x+\frac{1}{2}\sin ^{2}x-\frac{1}{3}\sin ^{3}x+% \frac{1}{4}\sin ^{4}x \\ &&+\sin x-\frac{1}{2}\sin ^{2}x+\frac{1}{3}\sin ^{3}x-\frac{1}{4}\sin ^{4}x) \\ &=&\sin x\left( \log (1+\sin x)-\sin x+\frac{1}{2}\sin ^{2}x-\frac{1}{3}\sin ^{3}x+\frac{1}{4}\sin ^{4}x\right) \\ &&+\sin ^{2}x-\frac{1}{2}\sin ^{3}x+\frac{1}{3}\sin ^{4}x-\frac{1}{4}\sin ^{5}x \\ &=&(x^{6})\left( \frac{\sin x}{x}\right) ^{6}\left( \frac{\log (1+\sin x)-\sin x+\frac{1}{2}\sin ^{2}x-\frac{1}{3}\sin ^{3}x+\frac{1}{4}\sin ^{4}x}{% \sin ^{5}x}\right) \\ &&+\sin ^{2}x-\frac{1}{2}\sin ^{3}x+\frac{1}{3}\sin ^{4}x-\frac{1}{4}\sin ^{5}x. \end{eqnarray*} Now let us write the resulting expression of $g(x)$ as follows \begin{eqnarray*} g(x) &=&\{\sin x-2(1+\cos x)\}A(x)-B(x)-C(x) \\ &=&\{\sin x-2(1+\cos x)\}\left( \frac{\log (1+(\cos x-1))-(\cos x-1)+\frac{1% }{2}(\cos x-1)^{2}}{(\cos x-1)^{3}}\right) \left( \frac{\cos x-1}{x^{2}}% \right) ^{3}\left( x^{6}\right) \\ &&+\{\sin x-2(1+\cos x)\}\{(\cos x-1)-\frac{1}{2}(\cos x-1)^{2}\} \\ &&-\left( x^{6}\right) \left( \frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}}\right) -x^{2}+\frac{1}{6}x^{4} \\ &&-(x^{6})\left( \frac{\sin x}{x}\right) ^{6}\left( \frac{\log (1+\sin x)-\sin x+\frac{1}{2}\sin ^{2}x-\frac{1}{3}\sin ^{3}x+\frac{1}{4}\sin ^{4}x}{% \sin ^{5}x}\right) \\ &&-\sin ^{2}x+\frac{1}{2}\sin ^{3}x-\frac{1}{3}\sin ^{4}x+\frac{1}{4}\sin ^{5}x. \end{eqnarray*} Divide $g(x)$ by $x^{6}$ it follows that \begin{eqnarray*} \frac{g(x)}{x^{6}} &=&\left( \sin x-2(1+\cos x)\right) \left( \frac{\log (1+(\cos x-1))-(\cos x-1)+\frac{1}{2}(\cos x-1)^{2}}{(\cos x-1)^{3}}\right) \left( \frac{\cos x-1}{x^{2}}\right) ^{3} \\ &&-\left( \frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}}\right) -\left( \frac{\sin x% }{x}\right) ^{6}\lim_{x\rightarrow 0}\left( \frac{\log (1+\sin x)-\sin x+% \frac{1}{2}\sin ^{2}x-\frac{1}{3}\sin ^{3}x+\frac{1}{4}\sin ^{4}x}{\sin ^{5}x% }\right) \\ &&+((\sin x-2(1+\cos x))((\cos x-1)-\frac{1}{2}(\cos x-1)^{2}) \\ &&-x^{2}+\frac{1}{6}x^{4}-\sin ^{2}x+\frac{1}{2}\sin ^{3}x-\frac{1}{3}\sin ^{4}x+\frac{1}{4}\sin ^{5}x)/x^{6}. \end{eqnarray*} Let \begin{eqnarray*} h(x) &=&\{\sin x-2(1+\cos x)\}\{(\cos x-1)-\frac{1}{2}(\cos x-1)^{2}\} \\ &&-x^{2}+\frac{1}{6}x^{4}-\sin ^{2}x+\frac{1}{2}\sin ^{3}x-\frac{1}{3}\sin ^{4}x+\frac{1}{4}\sin ^{5}x. \end{eqnarray*} It remains just to prove that \begin{equation*} \lim_{x\rightarrow 0}\frac{h(x)}{x^{6}}=\frac{7}{120} \end{equation*} which is an easy computation (with or without) using LHR.

${\bf UPDATE:}$ The purpose of the first steps previously done reduced the computation of the limit of $% \frac{g(x)}{x^{6}}$ which is a complicated expression to the computation of the limit of $\frac{h(x)}{x^{6}}$ which is very simple comparatively to $% \frac{g(x)}{x^{6}}.$ Indeed, one can use the l'Hospital's rule six times very easily, but before starting to do the derivations some trigonometric simplifications are used as follows. This is

\begin{equation*} h(x)=(\sin x-2(1+\cos x))((\cos x-1)-\frac{1}{2}(\cos x-1)^{2})-x^{2}+\frac{1% }{6}x^{4}-\sin ^{2}x+\frac{1}{2}\sin ^{3}x-\frac{1}{3}\sin ^{4}x+\frac{1}{4}% \sin ^{5}x. \end{equation*}

Develop the product of the first two parenthesis and next simplifying and using power-reduction formulas ( \begin{equation*} \begin{array}{ccc} \sin ^{2}\theta =\frac{1-\cos 2\theta }{2} & & \cos ^{2}\theta =\frac{% 1+\cos 2\theta }{2} \\ \sin ^{3}\theta =\frac{3\sin \theta -\sin 3\theta }{4} & & \cos ^{3}\theta =% \frac{3\cos \theta +\cos 3\theta }{4} \\ \sin ^{4}\theta =\frac{3-4\cos 2\theta +\cos 4\theta }{8} & & \cos ^{4}\theta =\frac{3+4\cos 2\theta +\cos 4\theta }{8} \\ \sin ^{5}\theta =\frac{10\sin \theta -5\sin 3\theta +\sin 5\theta }{16} & & \cos ^{5}\theta =\frac{10\cos \theta +5\cos 3\theta +\cos 5\theta }{16}% \end{array} \end{equation*}

one then can re-write $h(x)\ $as follows \begin{equation*} h(x)=\frac{1}{6}x^{4}-\frac{35}{32}\sin x-x^{2}-\frac{1}{4}\cos x-\frac{5}{6}% \cos 2x+\frac{1}{4}\cos 3x- \frac{1}{24}\cos 4x+\sin 2x-\frac{21}{% 64}\sin 3x+\frac{1}{64}\sin 5x+\frac{7}{8} \end{equation*}

and therefore derivatives are simply calculated (because there is no power on the top of sin and cos), so \begin{equation*} h^{\prime }(x)=\frac{1}{4}\sin x-\frac{35}{32}\cos x-2x+\frac{2}{3}% x^{3}+2\cos 2x-\frac{63}{64}\cos 3x+ \frac{5}{64}\cos 5x+\frac{5}{% 3}\sin 2x-\frac{3}{4}\sin 3x+\frac{1}{6}\sin 4x \end{equation*} \begin{equation*} h^{\prime \prime }(x)=\frac{1}{4}\cos x+\frac{35}{32}\sin x+2x^{2}+\frac{10}{% 3}\cos 2x-\frac{9}{4}\cos 3x+ \frac{2}{3}\cos 4x-4\sin 2x+\frac{% 189}{64}\sin 3x-\frac{25}{64}\sin 5x- 2 \end{equation*} \begin{equation*} h^{\prime \prime \prime }(x)=4x+\frac{35}{32}\cos x-\frac{1}{4}\sin x-8\cos 2x+\frac{567}{64}\cos 3x- \frac{125}{64}\cos 5x-\frac{20}{3}\sin 2x+\frac{27}{4}\sin 3x-\frac{8}{3}\sin 4x \end{equation*} \begin{equation*} h^{(4)}(x)=\frac{81}{4}\cos 3x-\frac{35}{32}\sin x-\frac{40}{3}\cos 2x-\frac{% 1}{4}\cos x-\frac{32}{3} \cos 4x+16\sin 2x-\frac{1701}{64}\sin 3x+% \frac{625}{64}\sin 5x+4 \end{equation*} \begin{equation*} h^{(5)}(x)=\frac{1}{4}\sin x-\frac{35}{32}\cos x+32\cos 2x-\frac{5103}{64}% \cos 3x+ \frac{3125}{64}\cos 5x+\frac{80}{3}\sin 2x-\frac{243}{4}% \sin 3x+\frac{128}{3}\sin 4x \end{equation*} \begin{equation*} h^{(6)}(x)=\frac{1}{4}\cos x+\frac{35}{32}\sin x+\frac{160}{3}\cos 2x-\frac{% 729}{4}\cos 3x+ \frac{512}{3}\cos 4x-64\sin 2x+\frac{15\,309}{64}% \sin 3x-\frac{15\,625}{64} \sin 5x \end{equation*}

\begin{eqnarray*} \lim_{x\rightarrow 0}\frac{h(x)}{x^{6}} &=&\lim_{x\rightarrow 0}\frac{% h^{\prime }(x)}{6x^{5}}=\cdots =\lim_{x\rightarrow 0}\frac{h^{(6)}(x)}{6!}=% \frac{h^{(6)}(0)}{6!} \\ &=&\frac{1}{6!}\left( \frac{1}{4}\cos (0)+\frac{160}{3}\cos (0)-\frac{729}{4}% \cos (0)+ \frac{512}{3}\cos (0)\right) \\ &=&\frac{1}{6!}\left( \frac{1}{4}+\frac{160}{3}-\frac{729}{4}+\frac{512}{3}% \right) \\ &=&\frac{7}{120}. \end{eqnarray*} By the way, one have to verify that at each level (except the last one) the current derivative is $zero$ for $x=0,$ in order to be able to re-use LHR.

share|cite|improve this answer
@paramanand. I have skipped some explanations related to standard limits because I believe they are evident for you. – Idris Jun 17 at 6:30
See an approach based on Taylor series in my answer. – Paramanand Singh Jun 17 at 10:55
The computation you have done are based on Taylor series expansions of $\sin x, \cos x, \log(1 + x)$ and are are bit too lengthy. Also the expression $h(x)$ at the end of your answer seems complicated and I believe calculation of limit of $h(x)/x^{6}$ is also a bit laborious. I am still looking for a cleaner solution. +1 for these heavy computations by the way. – Paramanand Singh Jun 18 at 3:56
@Paramanand The calculation of the limit of $h(x)/x^6$ by LHR are very easy. I can do them if you want. If you want them without LHR they will be somewhat lengthy but this due to the higher order $x^6$. For me, the heart of the solution is the first part I did yesterday because of the manipulations I did are not usual. – Idris Jun 18 at 4:02
@ParamanandSingh. Concerning the first step of yesterday, I mean the reduction of computation of limit of $g(x)/x^6$ to the computation of the limit of $h(x)/x^6$, it is possible to use Taylor series in the same fashion as I did my manipulations. Also, I hope this time you are convinced that the computation of the limit of $h(x)/x^6$ is simple (and lengthy because the higher order is involved $x^6$.) I am happy to have contributed in these calculations. In doing so, I am now able to compute a limit which is still opened, I have asked months ago it is about $(tan(sin(x))-sin(tan(x)))/x^7$. – Idris Jun 18 at 6:40

Here is my work for the first limit: Let $f(x) = \frac{num(x)}{den(x)}$. We get $\infty/\infty$ so we can use L'Hopital:

\begin{align*} \frac{num'(x)}{den'(x)}&=\frac{3(1+\sec(x))\log \sec x}{(\tan(x))(x + \log(\sec x + \tan x)) + (\log \sec (x)) (1 + \sec(x)) }\\ &= \frac{3(1+\sec(x))\log \sec x}{(\tan(x))[x + (\log \sec x) +\log(1+\sin(x))] + (\log \sec x)(1+\sec(x)) )}\\ &= \frac{3}{\left(\frac{\tan(x)}{1+\sec(x)}\right)\left(1 + \frac{x + \log(1 + \sin(x))}{\log \sec x}\right)+ 1} \end{align*} Taking a limit as $x\rightarrow \pi/2^-$ gives $3/2$.

share|cite|improve this answer
See an approach based on Taylor series in my answer. – Paramanand Singh Jun 17 at 10:54

While I am still searching for a simple solution based on LHR, I found that method of Taylor series can also be applied without much difficulty. However we will need to make the substitution $\tan x = t$ so that $\sec x = \sqrt{1 + t^{2}}$ and as $x \to 0$ we also have $t \to 0$. We can do some simplification as follows \begin{align} A &= \lim_{x \to 0}\frac{2(1 + \sec x)\log \sec x - \tan x\{x + \log(\sec x + \tan x)\}}{x^{6}}\notag\\ &= \lim_{x \to 0}\frac{(1 + \sec x)\log(1 + \tan^{2}x) - \tan x\{x + \log(\sec x + \tan x)\}}{\tan^{6}x}\cdot\frac{\tan^{6}x}{x^{6}}\notag\\ &= \lim_{t \to 0}\frac{(1 + \sqrt{1 + t^{2}})\log(1 + t^{2}) - t\{\tan^{-1}t + \log(t + \sqrt{1 + t^{2}})\}}{t^{6}}\notag\\ &= \lim_{t \to 0}\frac{g(t) - h(t)}{t^{6}}\notag \end{align} The Taylor series expansions of functions in the numerator are easy to find with the exception of $\log(t + \sqrt{1 + t^{2}})$. But then we know that \begin{align} \log(t + \sqrt{1 + t^{2}}) &= \int_{0}^{t}\frac{dx}{\sqrt{1 + x^{2}}}\notag\\ &= \int_{0}^{t}\left(1 - \frac{x^{2}}{2} + \frac{3x^{4}}{8} + o(x^{4})\right)\,dx\notag\\ &= t - \frac{t^{3}}{6} + \frac{3t^{5}}{40} + o(t^{5})\notag\\ \end{align} Thus we can see that \begin{align} h(t) &= t\{\tan^{-1}t + \log(t + \sqrt{1 + t^{2}})\}\notag\\ &= t\left(t - \frac{t^{3}}{3} + \frac{t^{5}}{5} + t - \frac{t^{3}}{6} + \frac{3t^{5}}{40} + o(t^{5})\right)\notag\\ &= 2t^{2} - \frac{t^{4}}{2} + \frac{11t^{6}}{40} + o(t^{6})\notag \end{align} And further \begin{align} g(t) &= (1 + \sqrt{1 + t^{2}})\log(1 + t^{2})\notag\\ &= \left(2 + \frac{t^{2}}{2} - \frac{t^{4}}{8} + o(t^{4})\right)\left(t^{2} - \frac{t^{4}}{2} + \frac{t^{6}}{3} + o(t^{6})\right)\notag\\ &= t^{2}\left(2 + \frac{t^{2}}{2} - \frac{t^{4}}{8} + o(t^{4})\right)\left(1 - \frac{t^{2}}{2} + \frac{t^{4}}{3} + o(t^{4})\right)\notag\\ &= t^{2}\left(2 - \frac{t^{2}}{2} + \frac{7t^{4}}{24} + o(t^{4})\right)\notag\\ &= 2t^{2} - \frac{t^{4}}{2} + \frac{7t^{6}}{24} + o(t^{6})\notag \end{align} It is now clear that $$A = \lim_{t \to 0}\frac{g(t) - h(t)}{t^{6}} = \frac{7}{24} - \frac{11}{40} = \frac{35 - 33}{120} = \frac{1}{60}$$ and hence $$L = \frac{A}{7} = \frac{1}{420}$$ Getting rid of trigonometric functions $\sec x, \tan x$ does help in having simpler Taylor series which require very little amount of calculation. The only trigonometric function is $\tan^{-1}t$ which has the simplest Taylor series.

Another perhaps simpler approach via Taylor series would be to use the series for $\sec x, \tan x$ followed by integration (to get series for $\log \sec x$, $\log(\sec x + \tan x)$) and thus finding a Taylor series for $f(x)$ directly leading to $$f(x) = 1 + \frac{x^{4}}{420} + o(x^{4})$$ Thus we can start with $$\tan x = x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + o(x^{5})$$ and $$\sec x = \dfrac{1}{\cos x} = \dfrac{1}{1 - \dfrac{x^{2}}{2} + \dfrac{x^{4}}{24} + o(x^{4})} = 1 + \frac{x^{2}}{2} + \frac{5x^{4}}{24} + o(x^{4})$$ and on integrating the series for $\tan x, \sec x$ we get \begin{align} \log \sec x &= \frac{x^{2}}{2} + \frac{x^{4}}{12} + \frac{x^{6}}{45} + o(x^{6})\notag\\ \log(\sec x + \tan x) &= x + \frac{x^{3}}{6} + \frac{x^{5}}{24} + o(x^{5})\notag\\ \Rightarrow (1 + \sec t)\log \sec t &= \left(2 + \frac{t^{2}}{2} + \frac{5t^{4}}{24} + o(t^{4})\right)\left(\frac{t^{2}}{2} + \frac{t^{4}}{12} + \frac{t^{6}}{45} + o(t^{6})\right)\notag\\ &= t^{2}\left(2 + \frac{t^{2}}{2} + \frac{5t^{4}}{24} + o(t^{4})\right)\left(\frac{1}{2} + \frac{t^{2}}{12} + \frac{t^{4}}{45} + o(t^{4})\right)\notag\\ &= t^{2}\left(1 + \frac{5t^{2}}{12} + \frac{137t^{4}}{720} + o(t^{4})\right)\notag\\ &= t^{2} + \frac{5t^{4}}{12} + \frac{137t^{6}}{720} + o(t^{6})\notag \end{align} It follows that $$a(x) = 3\int_{0}^{x}(1 + \sec t)\log\sec t\,dt = x^{3}\left(1 + \frac{x^{2}}{4} + \frac{137x^{4}}{1680} + o(x^{4})\right)$$ and \begin{align} b(x) &= \{x + \log(\sec x + \tan x)\}\log \sec x\notag\\ &= \left(2x + \frac{x^{3}}{6} + \frac{x^{5}}{24} + o(x^{5})\right)\left(\frac{x^{2}}{2} + \frac{x^{4}}{12} + \frac{x^{6}}{45} + o(x^{6})\right)\notag\\ &= x^{3}\left(2 + \frac{x^{2}}{6} + \frac{x^{4}}{24} + o(x^{4})\right)\left(\frac{1}{2} + \frac{x^{2}}{12} + \frac{x^{4}}{45} + o(x^{4})\right)\notag\\ &= x^{3}\left(1 + \frac{x^{2}}{4} + \frac{19x^{4}}{240} + o(x^{4})\right)\notag \end{align} Thus we can see that \begin{align} f(x) &= \frac{a(x)}{b(x)} = \dfrac{1 + \dfrac{x^{2}}{4} + \dfrac{137x^{4}}{1680} + o(x^{4})}{1 + \dfrac{x^{2}}{4} + \dfrac{19x^{4}}{240} + o(x^{4})}\notag\\ &= 1 + px^{2} + qx^{4} + o(x^{4})\notag \end{align} where $$p + \frac{1}{4} = \frac{1}{4},\, q + \frac{p}{4} + \frac{19}{240} = \frac{137}{1680}$$ so that $$p = 0, q = \frac{137 - 133}{1680} = \frac{4}{1680} = \frac{1}{420}$$ and finally we get $$f(x) = 1 + \frac{x^{4}}{420} + o(x^{4})$$ as desired. This however does involve the division of power series one time and multiplication of power series 2 times (not to mention the use of not so familiar Taylor series for $\sec x, \tan x$ in the first place). The first approach which I have given in my answer uses the product of two series only one time and one application of LHR (given in the question itself).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.