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Let $Gap(n)$ be the smallest natural number k , such that $ ({k,k+1,...,k+n-1}) $ contains no prime numbers. Such a k obviously exists - for instance $k = (n+1)!+2$ has the desired property. My question is this: does $Gap(n)$ tend to $(n+1)! + 2$ as n goes to infinity, i .e. does the limit

$$lim \frac{Gap(n)}{(n+1)! + 2}$$ exist and, if so, is it equal to 1?

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    Look here : https://en.wikipedia.org/wiki/Prime_gap – 2017-02-02
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    The known smallest prime gaps are much smaller than $n!$. Not sure, whether it has been proven that this continues forever. – 2017-02-02
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    https://arxiv.org/abs/1412.5029 Some recent research on this. – 2017-05-23

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You have defined ${\rm Gap}(n)$ as the smallest composite $k$ beginning a prime gap of size $n$. Currently there is no theorem establishing the true order of magnitude of ${\rm Gap}(n)$. However, we can use Wolf's conjecture in this form: $$ {\rm Gap}(n) = p(n)+1 \asymp \sqrt{n} \exp(\sqrt{n}) = o(n!). $$ (The MathWorld article uses $p(n)$, the smallest prime followed by a prime gap of size $n$.)

Then this limit is zero: $$ \lim_{n\to\infty} \frac{{\rm Gap}(n)}{(n+1)! + 2} = \lim_{n\to\infty} \frac{\sqrt{n} \exp(\sqrt{n})}{(n+1)! + 2} = 0. $$ So, to the best of our current knowledge, the limit is not equal to 1 as the question suggested.