A calculus problem by Akshat Sharda

$\displaystyle \large \lim_{x\rightarrow 0} \frac{\left(\sin ^{-1}x-\tan ^{-1}x\right)}{\ln \left(1+x^3\right)}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\sin ^{-1}x-\tan ^{-1}x\right)}{x^3}\cdot \frac{x^3}{\ln \left(1+x^3\right)}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)-\tan ^{-1}x\right)}{x^3}\cdot 1$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\tan ^{-1}\frac{\left(\frac{x}{\sqrt{1-x^2}}-x\right)}{\left(1+\frac{x^2}{\sqrt{1-x^2}}\right)}\right)}{x^3}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\tan ^{-1}\left(\frac{\left(x-x\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}+x^2\right)}\right)\right)}{x^3}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\tan ^{-1}\left(\frac{\left(x-x\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}+x^2\right)}\right)\right)}{\frac{\left(x-x\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}+x^2\right)}}\cdot \frac{\frac{\left(x-x\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}+x^2\right)}}{x^3}$

$\displaystyle \large = \lim_{x\rightarrow 0} 1\cdot \frac{\frac{\left(x-x\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}+x^2\right)}}{x^3}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\frac{\left(1-\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}+x^2\right)}\right)}{x^2}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\frac{\left(1-\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}+x^2\right)}\cdot \frac{\left(1+\sqrt{1-x^2}\right)}{\left(1+\sqrt{1-x^2}\right)}\right)}{x^2}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{\left(\frac{x^2}{\left(\sqrt{1-x^2}+x^2\right)\left(1+\sqrt{1-x^2}\right)}\right)}{x^2}$

$\displaystyle \large = \lim_{x\rightarrow 0} \frac{1}{\left(\sqrt{1-x^2}+x^2\right)\left(1+\sqrt{1-x^2}\right)}$

$\displaystyle \large = \frac{1}{\left(\sqrt{1-0^2}+0^2\right)\left(1+\sqrt{1-0^2}\right)} = \boxed{0.5}$

We use the Taylor series for each term around $0$ .

$\displaystyle\sin^{-1} x = x + \frac{x^3}{6} + \frac{3x^5}{40} + O(x^6)$

$\displaystyle \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} + O(x^6)$ , so that

$\displaystyle \sin^{-1} x - \tan^{-1} x = \frac{x^3}{2} + O(x^5)$ . Also ,

$\displaystyle ln(1+x^3) = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - .... = x^3 + O(x^6)$ . It follows that

$\displaystyle lim_{x \rightarrow 0} \frac{ \sin^{-1} x - \tan^{-1} x}{ln(1+x^3)} = lim_{x \rightarrow 0} \frac{\frac{x^3}{2} + O(x^5)}{x^3 + O(x^6)} = lim_{x \rightarrow 0} \frac{\frac{1}{2} + O(x^2)}{1 + O(x^3)} = \frac{\frac{1}{2} + 0}{1+0} = \boxed{\frac{1}{2}}$ .

Write the Taylor series expansion for all the functions around 0. Limit can be obtained as 1/2 easily

Well i did the long way. I Used L'Hospital's rule twice To get the answer 1/2