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

Let the given limit be (L).
Then (L = \lim_{x \to 2} \frac{\tan(x-2)}{x^2 + (k-2)x - 2k})
Let (x - 2 = t). Then as (x \to 2), (t \to 0).
Also, (x = t + 2).
Then
(L = \lim_{t \to 0} \frac{\tan t}{(t+2)^2 + (k-2)(t+2) - 2k})
(L = \lim_{t \to 0} \frac{\tan t}{t^2 + 4t + 4 + kt + 2k - 2t - 2k - 2k})
(L = \lim_{t \to 0} \frac{\tan t}{t^2 + (4-2+k)t - 2k + 4})
(L = \lim_{t \to 0} \frac{\tan t}{t^2 + (2+k)t + 4 - 2k})
Since the limit is of the form (\frac{0}{0}), we must have (4 - 2k = 0), so (k = 2).
Then
(L = \lim_{t \to 0} \frac{\tan t}{t^2 + 4t} = \lim_{t \to 0} \frac{\tan t}{t} \cdot \lim_{t \to 0} \frac{1}{t+4} = 1 \cdot \frac{1}{4} = \frac{1}{4})
However, the question states that (L = 5). There must be a mistake in the question or our understanding of it.
Let's use L'Hopital's rule:
(L = \lim_{x \to 2} \frac{\sec^2(x-2)}{2x + k - 2} = \frac{1}{4 + k - 2} = \frac{1}{k+2} = 5)
(k + 2 = \frac{1}{5})
(k = \frac{1}{5} - 2 = -\frac{9}{5})
This value of k is not in the options. Let's re-examine the problem statement and limit expression.