$$ \lim _ { n \rightarrow \infty } \frac { ( 2 n ) ! } { 2 ^ { 2 n } } = ? $$

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To evaluate the limit

$$
\lim_{n \rightarrow \infty} \frac{(2n)!}{2^{2n}},
$$

we can use Stirling's approximation, which states that

$$
n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n.
$$

Applying this to $$ (2n)! $$:

$$
(2n)! \sim \sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}.
$$

Now substituting this approximation into our limit:

$$
\frac{(2n)!}{2^{2n}} \sim \frac{\sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}}{2^{2n}}.
$$

This simplifies to:

$$
\frac{\sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}}{(2^2)^{n}} = \frac{\sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}}{4^n} = \sqrt{4 \pi n} \left( \frac{n}{e} \right)^{2n}.
$$

Now we can rewrite the limit:

$$
\lim_{n \rightarrow \infty} \sqrt{4 \pi n} \left( \frac{n}{e} \right)^{2n}.
$$

As $$ n $$ approaches infinity, the term $$ \left( \frac{n}{e} \right)^{2n} $$ grows much faster than $$ \sqrt{4 \pi n} $$, leading to:

$$
\lim_{n \rightarrow \infty} \frac{(2n)!}{2^{2n}} = \infty.
$$

Thus, the final result is:

$$
\lim_{n \rightarrow \infty} \frac{(2n)!}{2^{2n}} = \infty.
$$
The limit of $$ \frac{(2n)!}{2^{2n}} $$ as $$ n $$ approaches infinity is infinity.

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