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? 
 A: Hint: Write $y=\sqrt{x}$ and $b=\sqrt{a}$. Then you get:$${y-b+\sqrt{y^2-b^2}\over \sqrt{y^4-b^4}}={\sqrt{y-b}+\sqrt{y+b}\over \sqrt{(y+b)(y^2+b^2)}}$$ So the limit is $1\over \sqrt{2a}$.
A: 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 $x \to a$.
A: May be, you could make life a bit simpler using $x=y+a$ which makes
$$\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}=\frac{\sqrt{a+y}-\sqrt{a}+\sqrt{y}}{\sqrt{y (2 a+y)}}$$ Now, use the generalized binomial theorem or Taylor series around $y=0$ to get 
$$\sqrt{a+y}=\sqrt{a}+\frac{y}{2 \sqrt{a}}+O\left(y^2\right)$$
$${\sqrt{y (2 a+y)}}= \sqrt{2a} \sqrt{y}+O\left(y^{3/2}\right)$$ which make finally 
$$\frac{\sqrt{a+y}-\sqrt{a}+\sqrt{y}}{\sqrt{y (2 a+y)}}=\frac{1}{\sqrt{2a} }+\frac{\sqrt{y}}{2 \sqrt{2}\,a}+O\left(y^1\right)$$ which shows the limit and also how it is approached when $y\to 0$.
