If $$\lim_{x \to 2} \frac{\tan(x - 2)\{x^2 + (k+2)x - 2k\}}{x^2 - 4x + 4} = 5$$, then k is equal to:

We are given the limit:

$$\lim_{x \to 2} \frac{\tan(x - 2)\{x^2 + (k-2)x - 2k\}}{x^2 - 4x + 4} = 5$$

First, we simplify the denominator. Notice that $$x^2 - 4x + 4 = (x - 2)^2$$. So the expression becomes:

$$\lim_{x \to 2} \frac{\tan(x - 2) \cdot \{x^2 + (k-2)x - 2k\}}{(x - 2)^2}$$

Next, we factor the quadratic expression in the numerator: $$x^2 + (k-2)x - 2k$$. To factor it, we find two numbers that multiply to $$-2k$$ and add to $$k-2$$. The discriminant is $$(k-2)^2 - 4(1)(-2k) = k^2 - 4k + 4 + 8k = k^2 + 4k + 4 = (k+2)^2$$. The roots are:

$$\frac{-(k-2) \pm \sqrt{(k+2)^2}}{2} = \frac{-(k-2) \pm (k+2)}{2}$$

So the roots are:

$$\frac{-k + 2 + k + 2}{2} = \frac{4}{2} = 2 \quad \text{and} \quad \frac{-k + 2 - k - 2}{2} = \frac{-2k}{2} = -k$$

Thus, the quadratic factors as $$(x - 2)(x + k)$$. Substituting this, the expression becomes:

$$\lim_{x \to 2} \frac{\tan(x - 2) \cdot (x - 2)(x + k)}{(x - 2)^2} = \lim_{x \to 2} \frac{\tan(x - 2) (x + k)}{x - 2}$$

for $$x \neq 2$$.

We can rewrite this as:

$$\lim_{x \to 2} (x + k) \cdot \frac{\tan(x - 2)}{x - 2}$$

As $$x \to 2$$, $$x + k \to 2 + k$$. For the second part, let $$u = x - 2$$. As $$x \to 2$$, $$u \to 0$$, so:

$$\lim_{u \to 0} \frac{\tan u}{u} = 1$$

by the standard limit. Therefore, the entire limit is:

$$(2 + k) \cdot 1 = 2 + k$$

Given that this limit equals 5, we set up the equation:

$$2 + k = 5$$

Solving for $$k$$:

$$k = 5 - 2 = 3$$

Now, comparing with the options:

A. 0

B. 1

C. 2

D. 3

We see that $$k = 3$$ corresponds to Option D.

Hence, the correct answer is Option D.