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.
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.
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