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 $f$