To find the value of [tex]\( k \)[/tex] such that the limit exists for the given piecewise function, we need to ensure that the left-hand limit and the right-hand limit at [tex]\( x = 2 \)[/tex] are equal.
The function is defined as:
[tex]\[ f(x) = \begin{cases}
k^2 x^2, & \text{if } x > 2 \\
2xk - 1, & \text{if } x \leq 2
\end{cases} \][/tex]
We need to compute the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 2 from both the left and the right and set these equal to find [tex]\( k \)[/tex].
### Right-Hand Limit (as [tex]\( x \)[/tex] approaches 2 from the right)
For [tex]\( x > 2 \)[/tex], the function is:
[tex]\[ f(x) = k^2 x^2 \][/tex]
[tex]\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} k^2 x^2 = k^2 (2)^2 = 4k^2 \][/tex]
### Left-Hand Limit (as [tex]\( x \)[/tex] approaches 2 from the left)
For [tex]\( x \leq 2 \)[/tex], the function is:
[tex]\[ f(x) = 2xk - 1 \][/tex]
[tex]\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2xk - 1) = 2(2)k - 1 = 4k - 1 \][/tex]
For the limit to exist at [tex]\( x = 2 \)[/tex], these limits must be equal:
[tex]\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^-} f(x) \][/tex]
Setting the two limits equal, we get:
[tex]\[ 4k^2 = 4k - 1 \][/tex]
### Solving for [tex]\( k \)[/tex]
To solve for [tex]\( k \)[/tex], we rearrange the equation:
[tex]\[ 4k^2 - 4k + 1 = 0 \][/tex]
This is a quadratic equation, and we can solve it using the quadratic formula [tex]\( k = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a} \)[/tex], where [tex]\( a = 4 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 1 \)[/tex].
[tex]\[ k = \frac{{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}}{2 \cdot 4} \][/tex]
[tex]\[ k = \frac{{4 \pm \sqrt{16 - 16}}}{8} \][/tex]
[tex]\[ k = \frac{{4 \pm \sqrt{0}}}{8} \][/tex]
[tex]\[ k = \frac{4}{8} \][/tex]
[tex]\[ k = \frac{1}{2} \][/tex]
So, the value of [tex]\( k \)[/tex] that ensures the limit exists is:
[tex]\[ k = \frac{1}{2} \][/tex]
Therefore, [tex]\( k = \frac{1}{2} \)[/tex] is the solution.
