Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 46

Given the function
$f(x) = \left\{ \begin{array}{c} 4 - x^2 & \text{ if } & x \leq 2\\ x - 1 & \text{ if } & x > 2 \end{array} \right.$


a.) Find $\lim\limits_{x \to 2^-} f(x)$ and $\lim\limits_{x \to 2^+} f(x)$


For left hand limit

$\begin{equation} \begin{aligned} \lim\limits_{x \to 2^-} f(x) & = \lim\limits_{x \to 2^-} (4-x^2) \\ \lim\limits_{x \to 2^-}(4-x^2) & = \lim\limits_{x \to 2^-} 4 - \lim\limits_{x \to 2^-} x^2\\ \lim\limits_{x \to 2^-} (4-x^2) & = 4 - (-2)^2 \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \to 2^-} (4-x^2)= 0}$


For right hand limit


$\begin{equation} \begin{aligned} \lim\limits_{x \to 2^+} f(x) = & \lim\limits_{x \to 2^+} (x-1)\\ \lim\limits_{x \to 2^+} f(x) = & \lim\limits_{x \to 2^+} (x-1)\\ \lim\limits_{x \to 2^+} (x-1) =& 2 - 1\\ \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \to 2^+}(x-1) = 1}$

b.) Does $\lim\limits_{x \to 2} f(x)$ exists?
No, it does not exist because the left and right hand limits are different.

c.)Sketch the graph of $f$