Find the value of the following limit.
![\[ \lim_{x \to 0^+} \frac{1}{x \sqrt{x}} \left( a \arctan \frac{\sqrt{x}}{a} - b \arctan \frac{\sqrt{x}}{b} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-211e3dcf49680135df9589f3cc84e06b_l3.png "Rendered by QuickLaTeX.com")
To simplify the computations, we make the substitution 
. Then,
![\[ \lim_{x \to 0^+} \frac{1}{x \sqrt{x}} \left( a \arctan \frac{\sqrt{x}}{a} - b \arctan \frac{\sqrt{x}}{b} \right) = \lim_{x \to 0^+} \frac{1}{t^3} \left( a \arctan \frac{t}{a} - b \arctan \frac{t}{b} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-792112bf05818e65e74f79e085c5f1f6_l3.png "Rendered by QuickLaTeX.com")
Now, we apply L’Hopital’s rule three times,
![\begin{align*} \lim_{t \to 0^+} \frac{1}{t^3} \left( a \arctan \frac{t}{a} - b \arctan \frac{t}{b} \right) &= \lim_{t \to 0^+} \frac{1}{3t^2} \left( \frac{a^2}{a^2+t^2} - \frac{b^2}{b^2+t^2} \right) \\[9pt] &= \lim_{t \to 0^+} \frac{1}{6t} \left( \frac{-2a^2 t}{(a^2+t^2)^2} + \frac{2b^2 t}{(b^2+t^2)^2} \right) \\[9pt] &= \lim_{x \to 0^+} \frac{1}{6} \left( \frac{-2a^2(a^2+t^2) + 8a^2 t^2}{(a^2+t^2)^3} + \frac{2b^2(b^2+t^2) - 8b^2 t^2}{(b^2+t^2)^3} \right) \\[9pt] &= \frac{1}{6} \left( \frac{-2}{a^2} + \frac{2}{b^2} \right) \\[9pt] &= \frac{1}{3} \left( \frac{a^2 - b^2}{(ab)^2} \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-be8e91aed24ffe749e61afaab0b47198_l3.png "Rendered by QuickLaTeX.com")
Since this limit exists, the application of L’Hopital’s rule was justified.
Find the value of the following limit.
To simplify the computations, we make the substitution . Then,
Now, we apply L’Hopital’s rule three times,
Since this limit exists, the application of L’Hopital’s rule was justified.