## Precalculus (6th Edition) Blitzer

$$f(x)=\begin{cases}x+3, & x \lt 2 \\ x+3, & x \gt 2 \end{cases}$$ (Other answers are possible.)
Consider the function$$f(x)=\begin{cases}x+3, & x \lt 2 \\ x+3, & x \gt 2 \end{cases}.$$This function is not defined at $x=2$. Nonetheless, the limit at $x=2$ exists and equals: $$\lim_{x \to 2}f(x)=5.$$ Please note that in the definition of the limit, $\lim_{x \to a}f(x)$, the function $f$ does not need to be defined at $x=a$.