Finding the limit of a function . How can I calculate the following limit:
\begin{equation*}
\lim_{x \rightarrow a}
\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}
\end{equation*}
I feel that I should multiply by the conjugate, but which conjugate? 

Note that the above limit doesn't make sense for $x < a$, since the square root in the numerator has a negative argument, but I avoid writing this.
$$
\frac{\sqrt x - \sqrt a + \sqrt{x-a}}{\sqrt{x^2-a^2}} = \frac{1}{\sqrt{x+a}}\left(\frac{\sqrt x - \sqrt a}{\sqrt{x-a}}  + 1\right) 
$$
Finally, $\lim_{x \to a} \frac{\sqrt x - \sqrt a}{\sqrt{x-a}} = 0$. This is because we can write this as $\frac{\sqrt x - \sqrt a}{x-a} \times \frac{x-a}{\sqrt{x-a}}$. The limit of the first term exists, since it is the derivative of the square root function at $a$. The second term is just $\sqrt{x-a}$, whose limit is zero. So the limit of the product exists and is zero.
Finally,
$$ \bbox[yellow,5px,border:2px solid red]
{
\lim_{x \to a} \frac{1}{\sqrt{x+a}} \left(\frac{\sqrt x - \sqrt a}{\sqrt{x-a}} + 1\right) = \frac{1}{\sqrt{2a}}
}
$$ 
since $\sqrt{x+a} \to \sqrt{2a}$ as $x \to a$.
