To find the limit of the given expression, we can simplify the binomial coefficient first.

We have:
[tex]\binom{2x}{x + 8} = \frac{(2x)!}{(x+8)!(2x-(x+8))!} = \frac{(2x)!}{(x+8)!(x-8)!}[/tex]

Now let's simplify the expression further:
[tex]\frac{(2x)!}{(x+8)!(x-8)!} = \frac{2x(2x-1)(2x-2)...(x+1)x(x-1)(x-2)...(9)(8)(7)...(1)}{(x+8)(x+7)...(3)(2)(1)(x-8)(x-9)...(3)(2)(1)}[/tex]

Note that many terms will cancel out, leaving us with:
[tex]\frac{2x(2x-1)(2x-2)...(x+1)x}{(x+8)(x+7)...(9)(8)[/tex]

Now, let's take the limit as x approaches infinity. We have:
[tex]lim \binom{2x}{x + 8} {}^{ - 1x} = lim \frac{2x(2x-1)(2x-2)...(x+1)x}{(x+8)(x+7)...(9)(8)} = \infty[/tex]

Therefore, the limit of the given expression as x approaches infinity is infinity.