# Question #2849d

Nov 24, 2017

If the denominator is supposed to be $x \sqrt{x}$, you can use L'Hopital's Rule to get an answer equal to $\frac{{a}^{2} - {b}^{2}}{3 {a}^{2} {b}^{2}}$ (when $a \ne 0$ and $b \ne 0$).

#### Explanation:

The limit ${\lim}_{x \to 0} \frac{a \arctan \left(\frac{\sqrt{x}}{a}\right) - b \arctan \left(\frac{\sqrt{x}}{b}\right)}{x \sqrt{x}}$ is a $\frac{0}{0}$ indeterminate form, so L'Hopital's Rule can be used.

The derivative of the numerator is

$\frac{a}{1 + \frac{x}{a} ^ 2} \cdot \frac{1}{2 a \sqrt{x}} - \frac{b}{1 + \frac{x}{b} ^ 2} \cdot \frac{1}{2 b \sqrt{x}}$

$= {a}^{2} / \left(2 {a}^{2} \sqrt{x} + 2 x \sqrt{x}\right) - {b}^{2} / \left(2 {b}^{2} \sqrt{x} + 2 x \sqrt{x}\right)$

$= {a}^{2} / \left(2 \sqrt{x} \left({a}^{2} + x\right)\right) - {b}^{2} / \left(2 \sqrt{x} \left({b}^{2} + x\right)\right)$

$= \frac{{a}^{2} \cdot \left({b}^{2} + x\right) - {b}^{2} \cdot \left({a}^{2} + x\right)}{2 \sqrt{x} \left({a}^{2} + x\right) \left({b}^{2} + x\right)}$

$= \frac{\left({a}^{2} - {b}^{2}\right) x}{2 \sqrt{x} \left({a}^{2} + x\right) \left({b}^{2} + x\right)}$

$= \frac{\left({a}^{2} - {b}^{2}\right) \sqrt{x}}{2 \left({a}^{2} + x\right) \left({b}^{2} + x\right)}$.

The derivative of the denominator is $\frac{3}{2} \sqrt{x}$

Therefore,

${\lim}_{x \to 0} \frac{a \arctan \left(\frac{\sqrt{x}}{a}\right) - b \arctan \left(\frac{\sqrt{x}}{b}\right)}{x \sqrt{x}}$

$= {\lim}_{x \to 0} \frac{\frac{\left({a}^{2} - {b}^{2}\right) \sqrt{x}}{2 \left({a}^{2} + x\right) \left({b}^{2} + x\right)}}{\frac{3}{2} \sqrt{x}}$

$= {\lim}_{x \to 0} \frac{\left({a}^{2} - {b}^{2}\right)}{3 \left({a}^{2} + x\right) \left({b}^{2} + x\right)} = \frac{{a}^{2} - {b}^{2}}{3 {a}^{2} {b}^{2}}$ when $a \ne 0$ and $b \ne 0$.