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I have a stuck on the problem of L'Hospital's Rule,

$\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ which is in I.F. $\frac{0}{0}$

If we use the rule, we will have

$\lim\limits_{x\to 0} \frac{\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}}{\sec^2x-\cos x}$.

So, I think that I approach this problem in the wrong way.

Have you guy any idea?

Authawich Narissayaporn
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4 Answers4

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Alternatively, one may use standard Taylor expansions, as $x \to 0$, $$ \begin{align} \sin x&=x-\frac{x^3}{6}+o(x^4) \\\tan x&=x+\frac{x^3}{3}+o(x^4) \\\arctan x&=x-\frac{x^3}{3}+o(x^4) \\\arcsin x&=x+\frac{x^3}{6}+o(x^4) \end{align} $$ giving, as $x \to 0$, $$ \frac{\arctan x - \arcsin x}{\tan x - \sin x}= \frac{-\frac{x^3}{2}+o(x^4)}{\frac{x^3}{2}+o(x^4)}=-1+o(x) \to -1. $$

Olivier Oloa
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Let $\arcsin x = t$ so that $\sin t = x$ and $$\tan t = \dfrac{x}{\sqrt{1 - x^{2}}}$$ so that $$\arcsin x = t = \arctan\dfrac{x}{\sqrt{1 - x^{2}}}$$ and then we can proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\tan x - \sin x}\notag\\ &= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\sin x(1 - \cos x)}\cdot\cos x\notag\\ &= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\sin x(1 - \cos x)}\notag\\ &= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\sin^{3}x}\cdot(1 + \cos x)\notag\\ &= 2\lim_{x \to 0}\frac{\arctan x - \arcsin x}{x^{3}}\cdot\frac{x^{3}}{\sin^{3}x}\notag\\ &= 2\lim_{x \to 0}\dfrac{\arctan x - \arctan \dfrac{x}{\sqrt{1 - x^{2}}}}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\dfrac{\arctan \left(\dfrac{x - \dfrac{x}{\sqrt{1 - x^{2}}}}{1 + \dfrac{x^{2}}{\sqrt{1 - x^{2}}}}\right)}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\dfrac{\arctan \left(\dfrac{x(\sqrt{1 - x^{2}} - 1)}{\sqrt{1 - x^{2}} + x^{2}}\right)}{\dfrac{x(\sqrt{1 - x^{2}} - 1)}{\sqrt{1 - x^{2}} + x^{2}}}\cdot\dfrac{\dfrac{x(\sqrt{1 - x^{2}} - 1)}{\sqrt{1 - x^{2}} + x^{2}}}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\frac{\sqrt{1 - x^{2}} - 1}{x^{2}(\sqrt{1 - x^{2}} + x^{2})}\notag\\ &= 2\lim_{x \to 0}\frac{\sqrt{1 - x^{2}} - 1}{x^{2}}\notag\\ &= 2\lim_{x \to 0}-\frac{1}{\sqrt{1 - x^{2}} + 1}\notag\\ &= -2\cdot\frac{1}{2}\notag\\ &= -1\notag \end{align} In the above we have used the standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = 1 = \lim_{x \to 0}\frac{\arctan x}{x}$$ There is no need to use advanced tools like Taylor's series and L'Hospital's Rule.

Paramanand Singh
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  • If this problem is found in the section of Non-L'hospital, this way will be the best. Thanks. – Authawich Narissayaporn Oct 16 '16 at 16:53
  • @AuthawichNarissayaporn: Well, other people here have given reasonably good answers using L'Hospital or Taylor so I though to add a different and simpler approach. BTW when one is learning calculus, it is best to avoid the technique of L'Hospital's Rule because it severely hampers concept building (the technique is best suited for time constrained competitive exams and even then Taylor series approach beats it by a wide margin). – Paramanand Singh Oct 16 '16 at 17:00
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Hint:use the L'Hospital's Rule three times $$\lim\limits_{x\to 0} \frac{(\arctan x - \arcsin x)'''}{(\tan x - \sin x)'''}$$

E.H.E
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The first step is OK, but you should try and simplify things before going on.

The numerator is $$ \frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}= \frac{\sqrt{1-x^2}-1-x^2}{(1+x^2)\sqrt{1-x^2}}= \frac{-x^2(3+x^2)}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)} $$ The denominator is $$ \frac{1}{\cos^2x}-\cos x=\frac{1-\cos^3x}{\cos^2x}= \frac{(1-\cos x)(1+\cos x+\cos^2x)}{\cos^2x} $$ so you can rewrite your limit as $$ \lim_{x\to0} \frac{x^2}{1-\cos x} \frac{-(3+x^2)\cos^2x}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)(1+\cos x+\cos^2x)} $$ which is easy to compute (the second fraction is not “indeterminate”).

Of course, Taylor expansion is easier.

egreg
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