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.
Best Answer
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 (https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Multiple-angle_formulae) \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.