Solution
= Solution
For fixed real $a\ne0$, write the square root as $|a|\sqrt{1+x^2/a^2}$ and use the <binomial series> with exponent $-1/2$:
$$
\boxed{\frac{x}{\sqrt{x^2+a^2}}=\frac x{|a|}-\frac{x^3}{2|a|^3}+\frac{3x^5}{8|a|^5}+O(x^7).}
$$
The <absolute values> are necessary for negative $a$. This <Taylor series> is valid for $|x|<|a|$. If $a=0$, the quotient is $x/|x|$ for $x\ne0$; it has no continuous extension at zero and therefore \b[has no Taylor expansion about zero].