# Evaluation of $\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } \quad$ without using L'Hospital rule

My aim is to evaluate the following limit

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } \quad$

which evaluates to $\frac{1}{2}$ after using L'Hospital rule for rather $5$ steps, which of course is a tedious task. Could anyone suggest some shorter and elegant methods for evaluating this limit?

Thanks.

• Using $\arcsin'(x) = (1-x^2)^{-1/2}$ and $\arctan'(x) = (1+x^2)^{-1}$ and the generalized binomial theorem it's straight forward to expand these functions in a power-series. – Winther May 22 '18 at 11:47
• Thanks, I was unaware of the expansions of inverse trigonometric functions. – Swapnil Das May 22 '18 at 12:39
• btw to derive the expression for the derivatives above you can consider $\arctan(\tan(x)) = x$, take the derivative and express the result in terms of $y = \tan(x)$. Similar for other trigonometric functions. – Winther May 22 '18 at 12:43
• Write numerator as $-(x-\sin^{-1}x)+(x-\tan^{-1}x)$ and use math.stackexchange.com/questions/387333/… – lab bhattacharjee May 25 '18 at 18:36

Using Maclaurin series:

$$\arcsin x-\arctan x=x+\frac16x^3+\frac3{40}x^5+\ldots-\left(x-\frac13x^3+\frac15x^5-\ldots\right)=$$

$$=\frac12x^3+\mathcal O(x^5)$$

so

$$\lim_{x\to0}\frac{\arcsin x-\arctan x}{x^3}=\lim_{x\to0}\frac{\frac12x^3+\mathcal O(x^5)}{x^3}=\frac12$$

• The use of series expansion is equivalent to the use of l'Hopital rule. – Cesareo May 22 '18 at 13:06
• @CesarEo No, it's not...but it is very similar, true. Yet the OP did not want to avoid derivatives and L'Hospital, but rather to have a faster, simpler method to evaluate that limit. – DonAntonio May 22 '18 at 13:08

We have

$$\sin(\arcsin(x)-\arctan(x))=\frac{x(1-\sqrt{1-x^2})}{\sqrt{1+x^2}}$$

and

$$\lim_{x\to 0}\left(\frac{\arcsin(x)-\arctan(x)}{x^3}\right)\equiv \lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)}\right)$$

then

$$\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)} = \frac{x^3(1-\sqrt{1-x^2})}{x^2\sqrt{1+x^2}\sin(x^3)} = \left(\frac{x^3}{\sin(x^3)}\right)\frac{x^2}{x^2(1+\sqrt{1-x^2})\sqrt{1+x^2}}$$

then

$$\lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{x^3}\right)\equiv\lim_{x\to 0}\left(\frac{x^3}{\sin(x^3)}\right)\frac{1}{(1+\sqrt{1-x^2})\sqrt{1+x^2}} = \frac{1}{2}$$

• Very nice and novel argument. +1 I would have instead written $\arcsin x-\arctan x=\arcsin(\dots)$. – Paramanand Singh May 23 '18 at 1:50

This is easy!

Lets start by noting that $$arcsin(x)=x+x^3/6+3x^5/40+...$$ Next $$arctan(x)=x-x^3/3+x^5/5-...$$ When this is substituted in you limit equation above we have for the numerator: $$x^3/2-x^5/8 ...$$ When this is divided by $x^3$ and we pass to the limit we get the simple result of 1/2.

• This is same answer as above...why? – DonAntonio May 23 '18 at 9:46