Looks like standard limit form

By binomial theorem :

$\begin{aligned} (1-x^2)^n & = \sum_{k=0}^n (-1)^k \binom nk x^{2k} \\ \int_0^1 (1-x^2)^n dx & = \color{#3D99F6} \sum_{k=0}^n \frac {(-1)^k \binom nk}{2k+1} \\ \implies \color{#3D99F6} \omega_n & = \int_0^1 (1-x^2)^n dx & \small \color{#3D99F6} \text{Let }t = x^2 \implies dt = 2x \ dx \\ & = \int_0^1 \frac {t^{-\frac 12}(1-t)^n}2 dt \\ & = \frac {\text{B}\left(\frac 12, n+1\right)}2 & \small \color{#3D99F6} \text{where }\text{B}(\cdot) \text{ denotes the beta function.} \\ & = \frac {\Gamma \left(\frac 12\right) \Gamma(n+1)}{2 \Gamma \left(n+\frac 32\right)} & \small \color{#3D99F6} \text{where }\Gamma (\cdot) \text{ denotes the gamma function.} \\ & = \frac {2^n n!}{(2n+1)!!} \\ & = \frac {(2n)!!}{(2n+1)!!} < 1 \end{aligned}$

Since $\omega_n < 1$ , $\implies \displaystyle \lim_{n \to \infty} \sqrt[n] {\omega_n} = 0$ and by Stirling's formula $\displaystyle \lim_{n \to \infty} \frac {n!}{e^n} \sim \lim_{n \to \infty} \sqrt{2\pi n}\left(\frac n{e^2}\right)^n = \infty$ .

$\begin{aligned} L & = \lim_{n \to \infty} \left(1+\frac {\sqrt[n]{\omega_n}}{n!} \right)^\frac {n!}{e^n} & \small \color{#3D99F6} \text{A }1^\infty \text{ case} \\ & = \exp \left(\lim_{n \to \infty} \frac {\sqrt[n]{\omega_n}}{e^n}\right) & \small \color{#3D99F6} \implies \lim_{x \to \infty} f(x)^{g(x)} = e^{\lim_{x \to \infty} g(x)(f(x)-1)} \\ & = e^0 = \boxed 1 & \small \color{#3D99F6} \text{Since }\lim_{n \to \infty} \sqrt[n] {\omega_n} = 0 \end{aligned}$

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References:

• Beta function
• Gamma function
• $1^\infty$ -limit cases

We can solve $\omega_n$ in different ways here is one of my approach $\omega_n= \sum_{k=0}^{n}(-1)^k \binom{n}{k} \dfrac{1}{2k+1} = \sum_{k=0}^{n}(-1)^k \binom{n}{k} \int_0^1 x^{2k}\,dx \\ =\int_0^1 \left(\sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{2k}\right)\,dx=\int_0^1 (1-x^2)^n \,dx \\ =^{x^2 \to x} \int_0^1 (1-x)^n x^{-\frac{1}{2}}\,dx= \dfrac{1}{2}\left[\dfrac{\Gamma (n+1) }{\Gamma \left(n+\frac{3}{2}\right)}\Gamma\left(\frac{1}{2}\right)\right]\\ =\dfrac{1}{2}\left[ \dfrac{n!\sqrt {\pi}}{\dfrac{(2n+1)!!\sqrt \pi }{2^{n+1}}}\right]=\dfrac{2^n n! }{(2n+1)!!}= \dfrac{4^n (n!)^2}{(2n+1)!}$

Now we solve for $\Omega =\lim_{n\to \infty} \left(1+\frac{\sqrt[n]{\omega_n}}{n!}\right)^{\frac{n!}{e^n}} =\exp\left(\lim_{n\to \infty} \dfrac{\sqrt[n]{\omega_n}}{e^n}\right) \\=\exp\left(\lim_{n\to \infty} \dfrac{4}{\sqrt[n]{(2n+1)}e^n }\sqrt[n]{\frac{(n!)^2}{(2n)!}}\right)\\ \sim \small{\exp\left(\lim_{n\to \infty} \dfrac{4}{\sqrt[n]{2n+1}e^n}\dfrac{\sqrt[2n]{n\pi}}{4} \right) =\exp\left(\lim_{n\to \infty} \dfrac{\sqrt[2n]{\pi}}{e^n}\right)=e^0=1}$ Note for all $k\geq 1$ inductively we can show that $\ 1 \leq n^{\frac{1} {kn}}\leq n^{\frac{1}{n}}\\ \therefore \lim_{n\to \infty} 1 \leq \lim_{n\to \infty} n^{\frac{1}{kn}}\leq \lim_{n\to \infty} n^{\frac{1}{n}} =1$ Set $k=2$ and hence by Squeeze theorem $\lim_{n\to\infty} n^{\frac{1}{2n}}=1$