Let \( f(x)=\left\{\begin{array}{ll}x^{2}, & \text { for } x<2 \\ 3 x, & \text { for } x>2\end{array}\right. \). Find each limit. \( \begin{array}{ll}\text { (A), } \lim _{x \rightarrow 2^{+}} f(x) & \text { (B) } \lim _{x \rightarrow 2^{-}} f(x) \\ \text { (C) } \lim _{x \rightarrow 2} f(x) & \text { (D) } f(2)\end{array} \) \( \begin{array}{l}\text { A. Select the correct choice and, if necessary, fill in the answer box to complete your choice. } \\ \text { (A) }(x)=\square \text { (Type an integer or a decimal.) } \\ \text { B. The limit does not exist. }\end{array} \)
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The Deep Dive
To find these limits, we evaluate \( f(x) \) as \( x \) approaches 2 from the left and right. For \( \lim _{x \rightarrow 2^{+}} f(x) \) (part A), we use \( f(x) = 3x \) for \( x > 2 \). Evaluating this as \( x \) approaches 2 from the right gives us: \[ \lim _{x \rightarrow 2^{+}} f(x) = 3(2) = 6. \] For \( \lim_{x \rightarrow 2^{-}} f(x) \) (part B), we use \( f(x) = x^2 \) for \( x < 2 \). Evaluating this as \( x \) approaches 2 from the left gives us: \[ \lim _{x \rightarrow 2^{-}} f(x) = (2)^2 = 4. \] Now, for \( \lim_{x \rightarrow 2} f(x) \) (part C), we see that the left-hand limit (4) and the right-hand limit (6) are different, thus: \[ \lim_{x \rightarrow 2} f(x) \text{ does not exist.} \] Lastly, \( f(2) \) (part D) is not defined in the given piecewise function since there is no definition provided for \( f(2) \). To summarize: A. \( \lim_{x \rightarrow 2^{+}} f(x) = 6 \) B. The limit does not exist. C. \( \lim_{x \rightarrow 2^{-}} f(x) = 4 \) D. \( f(2) \) is not defined. So based on the limits and continuity, we conclude: (A) \( 6 \) (B) The limit does not exist.