About the $\lim_{h\to0}\frac{\text{arcsec}\left(\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}\right)}{h}$ I want to prove that
$$\lim_{h\to0}\frac{\text{arcsec}(x+h)-\text{arcsec}(x)}{h}=\frac{1}{|x|\sqrt{x^2-1}}.$$
without using definition of derivative and by the following method:
Because
$$\text{arcsec}(p)-\text{arcsec}(q)=\text{arcsec}\left(\frac{pq}{1+\sqrt{p^2-1}\sqrt{q^2-1}}\right),p,q>0\,\text{or}\,p,q<0$$
I arrived at the following limit
$$\lim_{h\to0}\frac{\text{arcsec}\left(\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}\right)}{h}\tag{*}$$
For evaluating recent limit I wrote it's as follows
$$\lim_{h\to0}\frac{\text{arcsec}\left(\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}\right)}{\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}}\times\lim_{h\to0}\frac{\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}}{h}$$
but it converts to $0\times\infty$.
Please, help me for evaluating $(*)$.
Also for $\arcsin(x)$ by using this method see here.
 A: A simple approach is to put $\text{arcsec}\, x = y$ so that $\sec y = x$ and $\text{arcsec}\, (x + h) = y + k$ so that $\sec(y + k) = x + h$. It should be clear that $k$ tends to zero with $h$. And then the desired limit is $\lim_{k \to 0}\dfrac{k}{\sec(y + k) - \sec y}$ which can be easily handled by changing $\sec$ into $1/\cos$. Some people may find it as essentially equivalent to the rule for differentiation of inverse function.
In that case another approach is to use the following relation $$\text{arcsec}\, x = \arccos \left(\frac{1}{x}\right) = \frac{\pi}{2} - \arcsin\left(\frac{1}{x}\right)$$ Using this we get
$\displaystyle \begin{aligned}L &= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{1}{x}\right) - \arcsin\left(\dfrac{1}{x + h}\right)}{h}\\
&= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{1}{x}\sqrt{1 - \dfrac{1}{(x + h)^{2}}} - \dfrac{1}{x + h}\sqrt{1 - \dfrac{1}{x^{2}}}\right)}{h}\\
&= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}\right)}{h}\\
&= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}\right)}{\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}}\cdot\dfrac{\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}}{h}\\
&= \lim_{h \to 0}1\cdot\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{xh(x + h)}\text{ (because }\lim_{y \to 0}\dfrac{\arcsin y}{y} = 1\text{)}\\
&= \lim_{h \to 0}\dfrac{(x + h)^{2} - x^{2}}{xh(x + h)\left\{\sqrt{(x + h)^{2} - 1} + \sqrt{x^{2} - 1}\right\}}\\
&= \lim_{h \to 0}\dfrac{h(2x + h)}{xh(x + h)\left\{\sqrt{(x + h)^{2} - 1} + \sqrt{x^{2} - 1}\right\}}\\
&= \frac{2x}{x\cdot x\cdot 2\sqrt{x^{2} - 1}} = \frac{1}{x\sqrt{x^{2} - 1}}\end{aligned}$
A: And idea: why won't you use the theorem about the derivative of the inverse function?
$$\color{red}{\text{For}\;\;y>0\iff\;-\frac\pi2<y<\frac\pi2}\;:\;\;y=\sec x\;\;,\;\;y'=\frac{\sin x}{\cos^2x}=\tan x\sec x\implies$$
$$x=\text{arcsec}\,y\implies (\text{arcsec}\,y)'=\frac1{\sec x\tan x}=\frac1{y\frac{\sqrt{1-\cos^2x}}{\cos x}}=\frac1{y\sqrt{\frac{1-\cos^2x}{\cos^2x}}}=$$
$$=\frac1{y\sqrt{y^2-1}}\;\ldots\ldots\text{ and we're done!}$$
