Let's solve the given limit problem:

The limit is:

$\lim_{{h \to 0}} \frac{{(x + h)^2 + 4(x + h) - x^2 - 4x}}{h}$

To evaluate this limit, we can start by expanding $(x + h)^2$ and $4(x + h)$ in the numerator.

1. Expand $(x + h)^2$: $(x + h)^2 = x^2 + 2xh + h^2$

2. Expand $4(x + h)$: $4(x + h) = 4x + 4h$

3. Substitute these expansions into the expression: $\frac{{x^2 + 2xh + h^2 + 4x + 4h - x^2 - 4x}}{h}$

4. Simplify the expression by canceling terms:

   • The $x^2$ and $-x^2$ terms cancel.
   • The $4x$ and $-4x$ terms cancel.

   This leaves: $\frac{{2xh + h^2 + 4h}}{h}$

5. Factor out $h$ in the numerator: $\frac{h(2x + h + 4)}{h}$

6. Cancel $h$ in the numerator and denominator (assuming $h \neq 0$): $2x + h + 4$

7. Take the limit as $h \to 0$: $\lim_{{h \to 0}} (2x + h + 4) = 2x + 4$

Thus, the answer is:

$\boxed{2x + 4}$

The correct answer is the option with $2x + 4$, not "nonexistent."

Would you like further explanation, or have any questions?

Here are 5 related questions that could help deepen understanding:

1. How can the limit definition be used to find derivatives in calculus?
2. Why is it important to factor out terms in limits involving $h$ to evaluate the limit as $h \to 0$?
3. How does this type of limit relate to the derivative of a function?
4. What would change if the expression involved higher powers of $h$ in the numerator?
5. How can we recognize when a limit truly does not exist versus when it simplifies to an answer?

Tip: When dealing with limits, always look for opportunities to factor and cancel terms to simplify expressions, especially when you have terms that could lead to an indeterminate form like $\frac{0}{0}$.