Question

Find the limit (if it exists). If it does not exist, explain why. $$ \lim _{x \rightarrow 2} f(x), \text { where } f(x)=\left\\{\begin{array}{ll} x^{2}-4 x+6, & x<2 \\ -x^{2}+4 x-2, & x \geq 2 \end{array}\right. $$

Short answer

The limit of the function as \(x\) approaches 2 does not exist.

Step-by-step solution

- Find the left-hand limit

The left-hand limit refers to the limit of the function when \(x\) approaches 2 from the left (i.e., \(x < 2\)). So, for this part of the function, use the rule for the limit of a polynomial (which is simply the polynomial function evaluated at the limit point) to get: \[ \lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} (x^{2} - 4x + 6) = (2^{2} - 4*2 + 6) = 2. \]

- Find the right-hand limit

Similarly, the right-hand limit is the limit of the function as \(x\) approaches 2 from the right (i.e., \(x > 2\)). Evaluating the second piece of the function at \(x = 2\), we get: \[ \lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^+} (-x^{2} + 4x - 2) = (-2^{2} + 4*2 - 2) = -2. \]

- Determine if the overall limit exists

The overall limit of the function as \(x\) approaches 2 will exist if and only if the left-hand and right-hand limits exist and are equal. But as we've found in steps 1 and 2, the left-hand limit is 2 while the right-hand limit is -2. Therefore, the overall limit of the function as \(x\) approaches 2 does not exist because the two one-sided limits aren't equal.