If `lim_(xto2) (tan(x-2).(x^(2)+(k-2)x-2k))/((x^(2)-4x+4))=5`, then find the value of k.

Explanation

To solve the limit problem \( \lim_{x \to 2} \frac{\tan(x-2) \cdot (x^2 + (k-2)x - 2k)}{x^2 - 4x + 4} = 5 \), we will follow these steps:

### Step 1: Simplify the Denominator
The denominator can be simplified as follows:
\[ x^2 - 4x + 4 = (x-2)^2 \]
Thus, we rewrite the limit:
\[ \lim_{x \to 2} \frac{\tan(x-2) \cdot (x^2 + (k-2)x - 2k)}{(x-2)^2} \]

### Step 2: Analyze the Limit
As \( x \to 2 \), \( \tan(x-2) \) approaches \( \tan(0) = 0 \) and \( (x-2)^2 \) also approaches \( 0 \). Therefore, we have a \( \frac{0}{0} \) indeterminate form.

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator:
\[ \lim_{x \to 2} \frac{\frac{d}{dx}[\tan(x-2) \cdot (x^2 + (k-2)x - 2k)]}{\frac{d}{dx}[(x-2)^2]} \]

### Step 4: Differentiate the Numerator
Using the product rule, the derivative of the numerator is:
\[ \frac{d}{dx}[\tan(x-2)] \cdot (x^2 + (k-2)x - 2k) + \tan(x-2) \cdot \frac{d}{dx}[x^2 + (k-2)x - 2k] \]
Calculating these derivatives:
1. The derivative of \( \tan(x-2) \) is \( \sec^2(x-2) \).
2. The derivative of \( x^2 + (k-2)x - 2k \) is \( 2x + (k-2) \).

Thus, the derivative of the numerator becomes:
\[ \sec^2(x-2) \cdot (x^2 + (k-2)x - 2k) + \tan(x-2) \cdot (2x + (k-2)) \]

### Step 5: Differentiate the Denominator
The derivative of the denominator \( (x-2)^2 \) is:
\[ 2(x-2) \]

### Step 6: Evaluate the Limit
Now we evaluate the limit:
\[ \lim_{x \to 2} \frac{\sec^2(0) \cdot (2^2 + (k-2) \cdot 2 - 2k)}{2(2-2)} \]
Since \( \sec^2(0) = 1 \), we have:
\[ \lim_{x \to 2} \frac{(4 + 2(k-2) - 2k)}{0} \]
This simplifies to:
\[ \lim_{x \to 2} \frac{(4 + 2k - 4 - 2k)}{0} = \lim_{x \to 2} \frac{0}{0} \]
We need to ensure that the numerator approaches 0 as \( x \to 2 \).

### Step 7: Set the Numerator to Zero
For the limit to equal 5, we need to set the numerator equal to zero:
\[ 4 + 2(k-2) - 2k = 0 \]
This simplifies to:
\[ 4 + 2k - 4 - 2k = 0 \implies 0 = 0 \]
This is always true, so we need to find \( k \) such that the limit equals 5.

### Step 8: Solve for \( k \)
We set:
\[ \lim_{x \to 2} \frac{(2)(2) + (k-2)(2) - 2k}{2} = 5 \]
Substituting \( x = 2 \):
\[ \frac{4 + 2(k-2) - 2k}{2} = 5 \]
This simplifies to:
\[ \frac{4 + 2k - 4 - 2k}{2} = 5 \implies \frac{0}{2} = 5 \]
This means we have to find \( k \) such that:
\[ 4 + k - 2 = 5 \implies k = 3 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{3} \]