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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.

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  • Are you allowed to use Taylor expansion of function ArcSec[a] when "a" approaches 1 ? – Claude Leibovici Nov 25 '13 at 12:50
  • @ClaudeLeibovici: No, because I think that for calculating Taylor expansion of ArcSec[a] we must know the derivative of ArcSec[a]. –  Nov 25 '13 at 12:53
  • You are perfectly right and I am stupid ! Sorry for that. – Claude Leibovici Nov 25 '13 at 13:09
  • @user95733, http://cims.nyu.edu/~kiryl/teaching/Calculus-I/classSect4.4.pdf and http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions#Differentiating_the_inverse_cosine_function give the solution indirectly. I am trying to produce one like I did in the other answer you have linked in this question. – lab bhattacharjee Nov 25 '13 at 15:39

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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

$\displaystyle \begin{aligned}L &= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{1}{x}\right) - \arcsin\left(\dfrac{1}{x + h}\right)}{h}\\ &= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{1}{x}\sqrt{1 - \dfrac{1}{(x + h)^{2}}} - \dfrac{1}{x + h}\sqrt{1 - \dfrac{1}{x^{2}}}\right)}{h}\\ &= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}\right)}{h}\\ &= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}\right)}{\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}}\cdot\dfrac{\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}}{h}\\ &= \lim_{h \to 0}1\cdot\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{xh(x + h)}\text{ (because }\lim_{y \to 0}\dfrac{\arcsin y}{y} = 1\text{)}\\ &= \lim_{h \to 0}\dfrac{(x + h)^{2} - x^{2}}{xh(x + h)\left\{\sqrt{(x + h)^{2} - 1} + \sqrt{x^{2} - 1}\right\}}\\ &= \lim_{h \to 0}\dfrac{h(2x + h)}{xh(x + h)\left\{\sqrt{(x + h)^{2} - 1} + \sqrt{x^{2} - 1}\right\}}\\ &= \frac{2x}{x\cdot x\cdot 2\sqrt{x^{2} - 1}} = \frac{1}{x\sqrt{x^{2} - 1}}\end{aligned}$

Paramanand Singh
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    Note that using a direct relation dealing with $\text{arcsec}$ function is not a good idea as we don't have any standard limit formula available to manipulate $\text{arcsec}$. It is better to transform the problem into one dealing with $\arcsin$ because it has got a standard limit formula $\lim_{y \to 0}\dfrac{\arcsin y}{y} = 1$. This is similar to the case of direct trigonometric functions where all the limits problems ultimately make use of $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$. – Paramanand Singh Nov 26 '13 at 19:37
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And idea: why won't you use the theorem about the derivative of the inverse function?

$$\color{red}{\text{For}\;\;y>0\iff\;-\frac\pi2<y<\frac\pi2}\;:\;\;y=\sec x\;\;,\;\;y'=\frac{\sin x}{\cos^2x}=\tan x\sec x\implies$$

$$x=\text{arcsec}\,y\implies (\text{arcsec}\,y)'=\frac1{\sec x\tan x}=\frac1{y\frac{\sqrt{1-\cos^2x}}{\cos x}}=\frac1{y\sqrt{\frac{1-\cos^2x}{\cos^2x}}}=$$

$$=\frac1{y\sqrt{y^2-1}}\;\ldots\ldots\text{ and we're done!}$$

DonAntonio
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  • I don't want to find $(\text{arcsec}y)'$. I know your method. I know that $()$ equals to $(\text{arcsec}x)'$. but I want to find $()$ without using this. –  Nov 25 '13 at 14:00
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    Well, good luck then with that ugly limit! – DonAntonio Nov 25 '13 at 14:03