## Calculus with Applications (10th Edition)

$$f(x)=\frac{x^{2}-4}{x-2}$$ There is a discontinuity when $x=2$. So $$\lim _{x \rightarrow 2} f(x)=2+2=4.$$
$$f(x)=\frac{x^{2}-4}{x-2}$$ This rational function is discontinuous wherever the denominator is zero. There is a discontinuity when $x=2$. Since for $x \neq 2$ $$\begin{split} f(x)&=\frac{x^{2}-4}{x-2} \\ &=\frac{(x-2)(x+2)}{(x-2)} \\ &=(x+2) \end{split}$$ So $$\lim _{x \rightarrow 2} f(x)=2+2=4.$$