$$\lim_{x \to 0} \frac{x\left(e^{(\sqrt{1+x^2+x^4}-1)/x} - 1\right)}{\sqrt{1+x^2+x^4} - 1}$$

1. Identify the Limit and Define a Substitution

We are given the limit:

$$\lim_{x \to 0} \frac{x \left( e^{\frac{\sqrt{1+x^2+x^4}-1}{x}} - 1 \right)}{\sqrt{1+x^2+x^4}-1}$$

Let's define a new variable for the exponent of $$e$$ :

$$y = \frac{\sqrt{1+x^2+x^4}-1}{x}$$

From this substitution, we can express the denominator of our original limit as:

$$x \cdot y = \sqrt{1+x^2+x^4}-1$$

2. Evaluate the Limit of the Exponent as

$$x \to 0$$

Before substituting back into the main equation, let's find what $$y$$

approaches as $$x$$ approaches $$0$$ :

$$\lim_{x \to 0} y = \lim_{x \to 0} \frac{\sqrt{1+x^2+x^4}-1}{x}$$

To evaluate this, we multiply the numerator and denominator by the conjugate of the numerator: 

$$\lim_{x \to 0} \frac{\left(\sqrt{1+x^2+x^4}-1\right)\left(\sqrt{1+x^2+x^4}+1\right)}{x\left(\sqrt{1+x^2+x^4}+1\right)}$$

Simplify the numerator using the difference of squares
$$(a-b)(a+b) = a^2 - b^2$$ :

$$\lim_{x \to 0} \frac{1+x^2+x^4-1}{x\left(\sqrt{1+x^2+x^4}+1\right)}$$

$$\lim_{x \to 0} \frac{x^2+x^4}{x\left(\sqrt{1+x^2+x^4}+1\right)}$$

Factor out $$x$$

from the numerator to cancel with the denominator: 

$$\lim_{x \to 0} \frac{x(x+x^3)}{\sqrt{1+x^2+x^4}+1}$$

Now, substitute $$x = 0$$ :

$$\frac{0+0}{\sqrt{1+0+0}+1} = \frac{0}{2} = 0$$

So, as $$x \to 0$$ , we have $$y \to 0$$.

3. Substitute and Evaluate the Main Limit

Now, substitute $$y$$ and $$x \cdot y$$

back into the original limit expression:

$$\lim_{x \to 0} \frac{x \left( e^y - 1 \right)}{x \cdot y}$$

Cancel the $$x$$ terms:

$$\lim_{y \to 0} \frac{e^y - 1}{y}$$

This is a standard fundamental limit in calculus:

$$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$$

Final Answer

The limit evaluates to $$1$$ , which corresponds to Option B.