Answer Happy

Explain what it means to say that x→9−lim​f(x)=3 and x→9+lim​f(x)=2. As x approaches 9,f(x) approaches 2 , but f(9)=3. As x approaches 9 from the left, f(x) approaches 3. As x approaches 9 from the right, f(x) approaches 2. As x approaches 9,f(x) approaches 3 , but f(9)=2. As x approaches 9 from the right, f(x) approaches 3. As x approaches 9 from the left, f(x) approaches 2. In this situation is it possible that x→9lim​f(x) exists? Explain. Yes, f(x) could have a hole at (9,3) and be defined such that f(9)=2. Yes, f(x) could have a hole at (9,2) and be defined such that f(9)=3. Yes, if f(x) has a vertical asymptote at x=9, it can be defined such that x→9−lim​f(x)=3, x→9+lim​f(x)=2, and x→9lim​f(x) exists. No, x→9lim​f(x) cannot exist if x→9−lim​f(x)=x→9+lim​f(x)