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How to show that $$ \lim_{x\to0}\frac{\Gamma(x)}{\psi(x)}=-1 $$ where $\psi(x)$ is the digamma function.

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Note that $\Gamma(x) = \frac{1}{x} + h(x)$ for some function $h(x)$ holomorphic near the origin. Then $$\frac{\Gamma(x)}{\psi(x)}=\frac{\Gamma(x)^2}{\Gamma'(x)}=\frac{\frac{1}{x^{2}}‌​+\frac{2}{x}h(x)+h(x)^2}{-\frac{1}{x^{2}}+h'(x)}.$$ The rest is now clear. –  sos440 Sep 4 '12 at 5:30

1 Answer 1

up vote 8 down vote accepted

Simply use recurrence relations: $$ \frac{\Gamma(x)}{\psi(x)} = \frac{x \cdot \Gamma(x)}{x \cdot \psi(x)} = \frac{\Gamma(x+1)}{x \cdot \psi(x+1) - 1} $$ Since $\Gamma(1) = 1$ and $\psi(1)$ is finite, the limit readily follows.

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