Continuous Function Math Example 2

Solution

1. 1
   Find discontinuities: denominator $= 0$ when $x = 2$. Check numerator: $x^2-4 = (x-2)(x+2)$, so numerator is also $0$ at $x=2$.
2. 2
   Simplify: $f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$ for $x \neq 2$. The limit $\lim_{x\to2}f(x) = 4$ exists, but $f(2)$ is undefined.
3. 3
   This is a removable discontinuity. Define $\tilde{f}(x) = \begin{cases}f(x) & x\neq2 \\ 4 & x=2\end{cases}$ to make it continuous everywhere.