Gam-Lam

Calculus Level 3

$\lim_{x\to 0}\Gamma(x) \ln^{1-x} (1-x)$

Notation: $\Gamma(\cdot)$ denotes the gamma function .

The answer is -1.

1 solution

Dwaipayan Shikari
Apr 11, 2021

$\displaystyle\lim_{x\rightarrow{0}}\Gamma{(x)} \log^{(1-x)}(1-x) =\phi$

Now Maclaurin series for $\Gamma{(1+x)} = 1-\gamma x +\dfrac{\Gamma''{(1)}}{2!}x^2 +\cdots \implies \Gamma{(x)} =\dfrac{1}{x} -\gamma+O(x)$ And $\log(1-x)≈-x$

So $\phi = \displaystyle \lim_{x\rightarrow{0}} (\dfrac{1}{x}-\gamma)(-x) ^{(1-x)}=\lim_{x\rightarrow{0}} -(1-\gamma x) = -1$

Clever insight with the expansion of the Gamma function!! Elegant;))

Yoihenba Laishram - 2 months ago

Gam-Lam

The answer is -1.

1 solution

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