Evaluate the limit from both sides:
Positive side
$$\lim_{x \to 0^+}\left(e^\frac{8}{x}+8x\right)^\frac{x}{2}$$
Piggybacking off of Tomi's suggestion, let $p = \left(e^\frac{8}{x}+8x\right)^\frac{x}{2}$ such that
$$\ln(p) = \frac{x}{2} \ln\left(e^\frac{8}{x}+8x\right)$$
As $x \to 0$, $8x \to 0$. However, $e^\frac{8}{x}$ blows up towards positive infinity. In our limit we can cancel out the 8x as it is a very small term and we are left with
$$\lim_{x\to 0^+}\left(\ln(p)\right) = \frac{x}{2} \ln\left(e^\frac{8}{x}\right) = \frac{x}{2}\frac{8}{x} = 4$$
$$\lim_{x\to 0^+}(p) = e^4$$
Negative side
We return again to $p$. For $x\to0$, we still have $8x\to0$, but now $e^{8/x}\to 0 $ much more rapidly than $8x$ because the exponent $\frac{8}{x}\to -\infty$ for $x\to 0^-$.
Thus we are left with
$$(8x)^\frac{x}{2} = 8^\frac{x}{2}x^\frac{x}{2}$$
This expression is continuous everywhere, but is complex for $x<0$ and real for $x\ge0$. Thankfully, because it is continuous everywhere, its positive limit at 0 is equal to its negative limit at 0, hence we can implement
$$\lim_{x\to0^+}8^\frac{x}{2}x^\frac{x}{2}$$
$$=\lim_{x\to0^+}\exp \left(\ln(8^\frac{x}{2}x^\frac{x}{2})\right)$$
$$=\exp \left(\lim_{x\to0^+}\ln(8^\frac{x}{2})+\lim_{x\to0^+}\ln(x^\frac{x}{2})\right)$$
$$=\exp \left(\lim_{x\to0^+}\frac{x}{2}\ln(8)+\lim_{x\to0^+}\frac{x}{2}\ln(x)\right)$$
It is easy to demonstrate that $$\lim_{x\to0^+}\frac{x}{2}\ln(8)=0$$
However for $$\lim_{x\to0^+}\frac{x}{2}\ln(x)$$ we apply L'Hopital's rule.
$$\lim_{x\to0^+}\frac{x}{2}\ln(x)$$
$$=\frac{1}{2}\lim_{x\to0^+}\frac{\ln(x)}{1/x}$$
$$=\frac{1}{2}\lim_{x\to0^+}\frac{1/x}{-1/x^2}$$
$$=\frac{1}{2}\lim_{x\to0^+}(-x)$$
$$=0$$
Plugging each result back in,
$$\lim_{x\to0^-}8^\frac{x}{2}x^\frac{x}{2} = \exp(0) = 1$$
Putting it together.
Because
$$\lim_{x\to0^+}\left(e^\frac{8}{x}+8x\right)^\frac{x}{2} = 1 \neq \lim_{x\to 0^+}\left(e^\frac{8}{x}+8x\right)^\frac{x}{2} = e^4$$ the limit
$$\lim_{x\to0}\left(e^\frac{8}{x}+8x\right)^\frac{x}{2}$$
does not exist.