I used Wolfram Alpha for this problem and it gives me a "Puiseux expansion": $$\sqrt{x}-\frac{3 \sqrt{\frac{1}{x}}}{8}-\frac{7}{128}\left(\frac{1}{x}\right)^{3 / 2}-\frac{9\left(\frac{1}{x}\right)^{5 / 2}}{1024}+O\left(\left(\frac{1}{x}\right)^{3}\right)$$ which is exactly what I need.
My only question is how one would be able to obtain this expansion manually.
Note that the expansion you gave is for $\Gamma(x)/\Gamma(x-1/2)$. Using the beta function and an appropriate change of integration variables, we obtain \begin{align*} &\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi }}\frac{{\Gamma \left( {\frac{x}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {\frac{{x + 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi }}B\left( {\frac{x}{2},\frac{1}{2}} \right) = \frac{{x - 1}}{{2\sqrt \pi }}\int_0^1 {\frac{{t^{x/2 - 1} }}{{\sqrt {1 - t} }}dt} \\ & = \frac{{x - 1}}{{\sqrt \pi }}\int_0^{ + \infty } {\frac{{e^{ - xs} }}{{\sqrt {1 - e^{ - 2s} } }}ds} = \frac{{x - 1}}{{\sqrt {2\pi } }}\int_0^{ + \infty } {e^{ - xs} s^{ - 1/2} \sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}} ds} . \end{align*} By the definition of the Nörlund numbers (cf. https://mathworld.wolfram.com/NorlundPolynomial.html), $$ \sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}} = \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}s^n } $$ for $|s|<\pi$. Thus, by Watson's lemma (cf. http://dlmf.nist.gov/2.3.ii), \begin{align*} \frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} & \sim \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\int_0^{ + \infty } {e^{ - xs} s^{n - 1/2} ds} } \\ & = \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\Gamma \left( {n + \frac{1}{2}} \right)\frac{1}{{x^{n + 1/2} }}} \\ & = \sqrt {\frac{x}{2}} + \sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} }}{{2\sqrt \pi n!}}\Gamma \left( {n + \frac{1}{2}} \right)\left( {\frac{{2n + 1}}{{n + 1}}B_{n + 1}^{(1/2)} + B_n^{(1/2)} } \right)\left(\frac{2}{{x}}\right)^{n+1/2}} \end{align*} as $x\to +\infty$. It is possible to simplify this result to $$ \frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} \sim \sqrt {\frac{x}{2}} + \sum\limits_{n = 0}^\infty { \binom{ 1/2}{n + 1}B_{n + 1}^{(3/2)} \left(\frac{2}{{x}}\right)^{n+1/2}} . $$ Note that this is a divergent asymptotic expansion. See http://dlmf.nist.gov/5.11.iii for more general results and references.
