# How do you find the limit as x approaches 2 of the piecewise function f(x)={(4-x^2,if x<=2),(x-1,if x>2):} ?

Since the left-hand limit and the right-hand limit at $x = 2$ do not match, the limit ${\lim}_{x \to 2} f \left(x\right)$ does not exist.
${\lim}_{x \to {2}^{-}} f \left(x\right) = {\lim}_{x \to {2}^{-}} \left(4 - {x}^{2}\right) = 4 - {\left(2\right)}^{2} = 0$
${\lim}_{x \to {2}^{+}} f \left(x\right) = {\lim}_{x \to {2}^{+}} \left(x - 1\right) = \left(2\right) - 1 = 1$
Since the above one-sided limits at $x = 2$ do not match, the limit at $x = 2$ does not exist.