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

The correct option is D $\frac{a^2-b^2}{3a^2{b}^2}$

limit is of $\frac{0}{0}$ form

now using L'hopital's rule: $\underset{x\to 0^{+}}{lim}\frac{1}{x\sqrt{x}}(a{\tan }^{-1}\frac{\sqrt{x}}{a}-b{\tan }^{-1}\frac{\sqrt{x}}{b})={lim}_{x\to 0^{+}}\frac{2}{3\sqrt{x}}[\left(\frac{a}{1+\frac{x}{a^2}}\right)\frac{1}{2a\sqrt{x}}-\left(\frac{b}{1+\frac{x}{b^2}}\right)\frac{1}{2b\sqrt{x}}]={lim}_{x\to 0^{+}}\frac{1}{3x}[\left(\frac{a^2}{x+a^2}\right)-\left(\frac{b^2}{x+b^2}\right)]$

Again using L'hopital's rule:

$=\underset{x\to 0^{+}}{lim}\frac{1}{3}[-\frac{a^2}{{(x+a^2)}^2}+\frac{b^2}{{(x+b^2)}^2}]=\underset{x\to 0^{+}}{lim}\frac{1}{3}[-\frac{1}{a^2}+\frac{1}{b^2}]=\frac{a^2-b^2}{3a^2{b}^2}$