Find the value of the following limit.
![\[ \lim_{x \to a^+} \frac{ \sqrt{x} - \sqrt{a} + \sqrt{x-a}}{\sqrt{x^2-a^2}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cd0cb6cbb7fad87764932430d119ed21_l3.png "Rendered by QuickLaTeX.com")
We compute as follows,
![\begin{align*} \lim_{x \to a^+} \frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a}}{\sqrt{x^2-a^2}} &= \lim_{x \to a^+} \left( \frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}} + \frac{\sqrt{x-a}}{\sqrt{(x+a)(x-a)}} \right) \\[9pt] &= \lim_{x \to a^+} \left( \frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}} \right) + \frac{1}{2\sqrt{a}} \\[9pt] &= \frac{1}{2 \sqrt{a}} + \lim_{x \to a^+} \frac{ \frac{1}{2 \sqrt{x}}}{\frac{x}{\sqrt{x^2-a^2}}} &(\text{L'Hopital})\\[9pt] &= \frac{1}{2\sqrt{a}} + \lim_{x \to a^+} \frac{\sqrt{x^2-a^2}}{2x^{\frac{3}{2}}} \\[9pt] &= \frac{1}{2\sqrt{a}} + 0 = \frac{1}{2\sqrt{a}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2f633d5979df0c9b47ae0dabc7837fac_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.
We compute as follows,
Since this limit exists, the application of L’Hopital’s rule was justified.
The 2 at the end should be under the square root.