## Precalculus (6th Edition) Blitzer

discontinuous at $x= 5$
Step 1. The first rational function may have a hole at $x=-1$. which is not on the curve because $x\lt -1$. Thus the function is continuous for $x\lt -1$ Step 2. Examining the first boundary $x=-1$, we have $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}\frac{x^2-1}{x+1}=\lim_{x\to-1^-}(x-1)=-2$ and $\lim_{x\to-1^+}f(x)=2(-1)=-2=f(-1)$ Thus, the function is continuous at $x= -1$ Step 3. Examining the second boundary $x=5$, we have $\lim_{x\to5^-}f(x)=2(5)=10=f(5)$ and $\lim_{x\to5^+}f(x)=3(5)-4=11\ne f(5)$ Thus, the function is discontinuous at $x= 5$