Question

If $\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}\tan \left(\sin \left({\tan }^{-1}x\right)\right)+2\sqrt{1-x^2}\sin \left(\tan \left({\sin }^{-1}x\right)\right)-3x}{x^p}$ exists finitely, then which of the following is/are CORRECT?

• A. The value of $p$ is $3$
• B. The value of $p$ is $5$
• C. The value of limit is $\frac{-31}{60}$
• D. The value of limit is $\frac{-8}{5}$

Answer 1

Answer: (B), (C)

The value of limit is $\frac{-31}{60}$

$\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}\tan \sin {\tan }^{-1}x+2\sqrt{1-x^2}\sin \tan {\sin }^{-1}x-3x}{x^p}$
$\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}\tan \left(\frac{x}{\sqrt{1+x^2}}\right)+2\sqrt{1-x^2}\sin \left(\frac{x}{\sqrt{1-x^2}}\right)-3x}{x^p}$

Using expansion,
$\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}(\frac{x}{\sqrt{1+x^2}}+\frac{x^3}{3(1+x^2{)}^{3/2}}+\frac{2}{15}\frac{x^5}{(1+x^2{)}^{5/2}}+\dots )+2\sqrt{1-x^2}(\frac{x}{\sqrt{1-x^2}}-\frac{x^3}{3!(1-x^2{)}^{3/2}}+\frac{x^5}{5!(1-x^2)5/2}+\dots )-3x}{x^p}$

$\underset{x\to 0}{lim}\frac{(-\frac{2}{3(1-x^4)}+\frac{2}{15}\frac{1}{(1+x^2{)}^2}+\frac{2}{5!(1-x^2{)}^2})x^5}{x^p}$
$\Rightarrow p=5$ and limit $=-\frac{2}{3}+\frac{2}{15}+\frac{2}{5!}=\frac{-31}{60}$