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Prove that $$\lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}=\frac{1}{2}$$

$$ \lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}=\lim_{x→0} \frac{\tan^{-1}\frac{x}{\sqrt{1-x^2}}-\tan^{-1}x}{x^3} \\ =\lim_{x→0} \frac{\tan^{-1}\bigg[\frac{\frac{x}{\sqrt{1-x^2}}-x}{1+\frac{x^2}{\sqrt{1-x^2}}}\bigg]}{x^3} =\lim_{x→0} \frac{\tan^{-1}\bigg[\frac{x(1-\sqrt{1-x^2})}{\sqrt{1-x^2}+x^2}\bigg]}{x^3}\\ \text{since, }\frac{x^2}{\sqrt{1-x^2}}>-1 $$

I think I am getting stuck here, do not really see how to proceed further with the steps.

Note: I don't want to use L'Hospital's rule as I 'd like to solve it without using the derivatives of $\sin^{-1}x$ and $\tan^{-1}x$.

Sooraj S
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4 Answers4

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\begin{align*} &\lim_{x\rightarrow 0}\dfrac{\sin^{-1}x-\tan^{-1}x}{x^{3}}\\ &=\lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x-\tan^{-1}x)}{x^{3}}\cdot\dfrac{\sin^{-1}x-\tan^{-1}x}{\sin(\sin^{-1}x-\tan^{-1}x)}\\ &=\lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x-\tan^{-1}x)}{x^{3}}\cdot 1\\ &=\lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x)\cos(\tan^{-1}x)-\cos(\sin^{-1}x)\sin(\tan^{-1}x)}{x^{3}}\\ &=\lim_{x\rightarrow 0}\dfrac{x\cdot\dfrac{1}{\sqrt{1+x^{2}}}-\sqrt{1-x^{2}}\cdot\dfrac{x}{\sqrt{1+x^{2}}}}{x^{3}}\\ &=\lim_{x\rightarrow 0}\dfrac{1}{\sqrt{1+x^{2}}}\cdot\dfrac{1-\sqrt{1-x^{2}}}{x^{2}}\\ &=\lim_{x\rightarrow 0}\dfrac{1-(1-x^{2})}{x^{2}}\cdot\dfrac{1}{1+\sqrt{1-x^{2}}}\\ &=\lim_{x\rightarrow 0}\dfrac{1}{1+\sqrt{1-x^{2}}}\\ &=\dfrac{1}{2}. \end{align*}

user284331
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  • Personally I don't like this notation$$ \lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x-\tan^{-1}x)}{x^{3}}\cdot\dfrac{\sin^{-1}x-\tan^{-1}x}{\sin(\sin^{-1}x-\tan^{-1}x)}\ =\lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x-\tan^{-1}x)}{x^{3}}\cdot 1$$ – user Nov 25 '19 at 19:41
  • Sorry maybe it is trivial but I don't get this step $$\lim_{x\rightarrow 0}\dfrac{1}{\sqrt{1+x^{2}}}\cdot\dfrac{1-\sqrt{1-x^{2}}}{x^{2}}\ =\lim_{x\rightarrow 0}\dfrac{1-(1-x^{2})}{x^{2}}\cdot\dfrac{1}{1+\sqrt{1-x^{2}}}$$ – user Nov 25 '19 at 19:42
  • The $1/\sqrt{1+x^{2}}$ goes to $1$, so just omit it. For the rest, multiply it with the conjugate $\dfrac{1+\sqrt{1-x^{2}}}{1+\sqrt{1-x^{2}}}$. – user284331 Nov 25 '19 at 19:45
  • Ah ok sorry, I din't get the $\dfrac{1}{\sqrt{1+x^{2}}}$ term but it is not essential. This seems a very good way to prove this limit. – user Nov 25 '19 at 19:47
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Hint:

You can get rid of the $\arctan$ functions as

$$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}.$$ If $a$ and $b$ tend to zero, the denominator can be ignored, and

$$\frac{\tan(a-b)}{a-b}$$ tends to $1$.

Then

$$\frac x{x^3\sqrt{1-x^2}}-\frac x{x^3}=\frac 1{x^2\sqrt{1-x^2}}(1-\sqrt{1-x^2})=\frac1{\sqrt{1-x^2}(1+\sqrt{1-x^2})}.$$

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If you are not allowed to use Taylor’s series, we can assume that the limits as $x\to 0$

$$\frac{\tan^{-1}x-x}{x^3}=L_1\quad \frac{\sin^{-1}x-x}{x^3}=L_2$$

exists and show by algebraic manipulation that they are equal to $L_1=-\frac13$ and $L_2=\frac16$.

Then we can use these results to find the limit, indeed

$$\frac{\sin^{-1}x-\tan^{-1}x}{x^3}=\frac{\sin^{-1}x-x}{x^3}-\frac{\tan^{-1}x-x}{x^3} \to \frac16+\frac13=\frac12$$

Refer also to

user
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-2

hint

Sorry, i understood that $\sin^{-1}(x)$ means $\frac{1}{\sin(x)}$ instead of $\arcsin(x)$.

hamam_Abdallah
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