Let's do some calculus! (54)

We can see the limit is of the form $1^{\infty}$ .

$\displaystyle{\implies \lim_{n \rightarrow 0} \sqrt[n]{n!}}$

$\displaystyle{\implies \lim_{n \rightarrow 0} \sqrt[n]{\Gamma({n+1})}}$

As it is $1^{\infty}$ form , we apply,

$\displaystyle{\implies e^{\displaystyle{\lim_{n \rightarrow 0} \dfrac{\Gamma({n+1})-1}{n}}}}$

Now we can see the power is of the form $\dfrac{0}{0}$ , So we can Apply L'Hospital's rule ,

$\implies e^{\displaystyle{\lim_{n \rightarrow 0} (\Gamma'({n+1}))}}$

Now we know $\Gamma'(1)= - \gamma$ where $\gamma$ is Euler Mascheroni Const.

So we get Limit = $\boxed{e^{-\gamma}} \approx 0.561$

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Derivation of $\Gamma'(1)= -\gamma$ .

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To evaluate ${\Gamma}^{'} (1)$ , we set $n=1$ thus $\int _{ 0 }^{ \infty }{ { e }^{ -t }ln\left(t\right)} dt .$

We replace ${e}^{-t}$ with $\lim _{ n\rightarrow \infty }{ \left(1-\frac{t}{n} \right)^{n} } .$

Let $s = 1 - \frac{t}{n}$ and $-nds = dt,$ we get $\begin{aligned} \lim _{ n\rightarrow \infty }{ \left[ \int _{ 1}^{ 0}{ { s }^{ n } } \left[ ln(n)+ln(1-s) \right] (-n)ds \right] } \\ &= \lim _{ n\rightarrow \infty }{ \left[ nln(n)\int _{ 0 }^{ 1 }{ { s }^{ n } } ds+\int _{ 0 }^{ 1 }{ { s }^{ n } } ln(1-s)ds \right] } \\ &=\lim _{ n\rightarrow \infty }{ \left[ \frac { n }{ n+1 } ln(n)+\frac { -1 }{ n+1 } \int _{ 0 }^{ 1 }{ \frac { { s }^{ n+1 }-1 }{ s-1 } } ds \right] } \\ &=\lim _{ n\rightarrow \infty }{\frac { n }{ n+1 } \left[ ln(n)-{ H }_{ n+1 } \right]} . \end{aligned}$ ${H}_{n+1}$ is the harmonic number. By definition $\lim_{n\rightarrow \infty}{\left[{ H }_{ n+1 } -ln(n)\right]}$ is the Euler-Mascheroni constant; therefore ${\Gamma}^{'} (1) = -\gamma$ .