Question

$\underset{x\to 0^{+}}{lim}\frac{1}{x\sqrt{x}}(a{\tan }^{-1}\frac{\sqrt{x}}{a}-b{\tan }^{-1}\frac{\sqrt{x}}{b})$ has the value equal to

• A. $\frac{a-b}{3}$
• B. 0
• C. $\frac{(a^2-b^2)}{6a^2{b}^2}$
• D. $\frac{a^2-b^2}{3a^2{b}^2}$

Answer 1

Answer: (D)

$\frac{a^2-b^2}{3a^2{b}^2}$

$=\underset{x\to 0^{+}}{lim}\frac{1}{x\sqrt{x}}(ata{n}^{-1}\frac{\sqrt{x}}{a}-bta{n}^{-1}\frac{\sqrt{x}}{b})$
$=\underset{x\to 0}{lim}\frac{a(\frac{\sqrt{x}}{a}-\frac{\sqrt{(x{)}^3}}{3a^3}+\frac{\sqrt{(x{)}^6}}{5a^6}+....)-b(\frac{\sqrt{x}}{b}-\frac{\sqrt{(x{)}^3}}{3b^3}+\frac{\sqrt{(x{)}^6}}{5a^6}+....)}{x\sqrt{x}}$
$\Rightarrow \frac{1}{3}$($\frac{1}{b^2}-\frac{1}{a^2})$ = $\frac{a^2-b^2}{3a^2{b}^2}$