Practice Limit

$\begin{aligned} L & = \lim_{n \to \infty} ({\color{#3D99F6}n!})^{\frac 1{n^2}} & \small \color{#3D99F6} \text{By Stirling's formula: }n! \sim \sqrt{2\pi n} \left(\frac ne \right)^n \\ & = \lim_{n \to \infty} \left({\color{#3D99F6}\sqrt{2\pi n} \left(\frac ne \right)^n}\right)^{\frac 1{n^2}} \\ & = \lim_{n \to \infty} (2\pi)^{\frac 1{2n^2}} e^{-\frac 1n} n^{\frac 1n + \frac 1{2n^2}} \\ & = \lim_{n \to \infty} n^{\frac 1n + \frac 1{2n^2}} \\ & = \lim_{n \to \infty} \exp \left(\left(\frac 1n + \frac 1{2n^2}\right) \ln n\right) \\ & = \exp \left(\lim_{n \to \infty} \left(\frac {\ln n}n + \frac {\ln n}{2n^2} \right) \right) & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case L'Hôpital's rule applies} \\ & = \exp \left(\lim_{n \to \infty} \left(\frac {\frac 1n}1 + \frac {\frac 1n}{4n} \right) \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }n \\ & = e^0 = \boxed 1 \end{aligned}$

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References :

• L'Hôpital's rule
• Stirling's formula

Stirling's approximation is $\log{n!}=n\log{n}-n+O(n)$ ; so $\log{n!^{\frac{1}{n^2}}}=\frac{1}{n} \log{n}-\frac{1}{n}+O \left( \frac{1}{n} \right) \to 0$ as $n\to \infty$ ; hence the limit is $\boxed1$ .

Let $y=\lim _{n \to \infty}\Large(n!)^{\large {\frac{1}{n^2}}}$ Therefore $\ln y=\lim _{n \to \infty}\frac{1}{n^2}\sum_{r=0}^{n-1}\ln(n-r)$ $\ln y=\lim _{n \to \infty}\frac{\ln n}{n^2} - \frac1{n}\sum_{r=0}^{n-1}\frac1n\ln(1-\frac{r}{n})$ $\ln y=0- \lim _{n \to \infty}\frac1{n}\int_{0}^{1}\ln(1-x)dx$ $\ln y=0- \lim _{n \to \infty}\frac1{n}(-1)$ $\ln y=0- 0$ $y=e^0=\boxed{1}$

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Alternate: $\Large(n!)^{\large {\frac{1}{n^2}}} \leq (n^n)^{\frac{1}{n^2}} \leq n^{\frac1 n}\to 1$

$\frac { 1 }{ { n }^{ 2 } }$ tends clearly to 0 (for $n$ grows up to infinity). And ${ anythingElseThanZero }^{ 0 }=1$ . Thus the answer is 1.

Is this way of thinking wrong, and fortunatly leads to the correct answer in this case? And if it is wrong, why ? I'm a little bit confused, because all answers are way more complicated than this one.