Question:

Step 1: Analyze the limit form The given limit is $$ \lim_{x \to 2} \frac{\tan{(x - 2)}{(x^2 + (k - 2)x - 2k)}}{x^2 - 4x + 4} = 5 $$ As $x \to 2$, the denominator $x^2 - 4x + 4 = (x-2)^2 \to 0$. For the limit to be a finite value, the numerator must also approach 0. This implies that the term $x^2 + (k-2)x - 2k$ must have $(x-2)$ as a factor to allow for cancellation with the denominator. Step 2: Factor the quadratic term in the numerator For $x^2 + (k-2)x - 2k$ to have $(x-2)$ as a factor, $x=2$ must be a root of the quadratic. Substituting $x=2$ into the quadratic expression: $$ 2^2 + (k-2)(2) - 2k = 4 + 2k - 4 - 2k = 0 $$ This equation is satisfied for all values of $k$, confirming that $(x-2)$ is always a factor of $x^2 + (k-2)x - 2k$. To find the other factor, we can use polynomial division or observe that if $x=2$ is a root, and the constant term is $-2k$, then the other root must be $-k$. Thus, the quadratic can be factored as: $$ x^2 + (k-2)x - 2k = (x-2)(x+k) $$ Step 3: Rewrite the limit expression Substitute the factored quadratic and the factored denominator into the limit expression: $$ \lim_{x \to 2} \frac{\tan(x-2) \cdot (x-2)(x+k)}{(x-2)^2} $$ For $x \neq 2$, we can simplify the expression by canceling one factor of $(x-2)$: $$ \lim_{x \to 2} \frac{\tan(x-2) \cdot (x+k)}{x-2} $$ This can be rearranged to separate the standard limit form: $$ \lim_{x \to 2} \left( \frac{\tan(x-2)}{x-2} \cdot (x+k) \right) $$ Step 4: Apply standard limit properties We use the known standard limit $\lim_{u \to 0} \frac{\tan u}{u} = 1$. Let $u = x-2$. As $x \to 2$, $u \to 0$. Therefore, $$ \lim_{x \to 2} \frac{\tan(x-2)}{x-2} = 1 $$ Step 5: Evaluate the limit and solve for k Substitute the value of the standard limit back into the expression: $$ 1 \cdot \lim_{x \to 2} (x+k) = 5 $$ Now, evaluate the remaining limit by direct substitution: $$ 1 \cdot (2+k) = 5 $$ $$ 2+k = 5 $$ Solving for $k$: $$ k = 3 $$