The problem shown in the image involves solving a limit:

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

Let's go step by step through the solution:

1. Rewrite the numerator: $(x - 2)^2 - 4 = (x^2 - 4x + 4) - 4$ This simplifies to: $x^2 - 4x$

2. Substitute this into the limit expression: $\lim_{{x \to 0}} \frac{{x^2 - 4x}}{x}$

3. Factor the numerator: $\frac{{x(x - 4)}}{x}$

4. Simplify by canceling the $x$: $x - 4$

5. Now, substitute $x = 0$ into the simplified expression: $0 - 4 = -4$

Therefore, the solution is:

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

Would you like more details on any of the steps? Here are 5 related questions:

1. What happens if you don't factor the numerator?
2. Why does canceling $x$ in the limit require careful attention?
3. How would the solution change if the limit was $x \to 2$?
4. What if the numerator was a cubic expression?
5. How can you apply L'Hôpital's Rule to this limit?

Tip: Always simplify the expression as much as possible before substituting into a limit to avoid indeterminate forms like $\frac{0}{0}$.