## Calculus: Early Transcendentals 8th Edition

The function is discontinuous because $f(3) = 6 \ne 7$.
For this function to be continuous by definition: $\lim\limits_{x \to 3} f(x) = f(3)$ $\lim\limits_{x \to 3} \frac{2x^{2}-5x-3}{x-3} =$ $\lim\limits_{x \to 3} \frac{(2x+1)(x-3)}{x-3} =$ Cancel out $(x-3)$ $\lim\limits_{x \to 3} (2x+1) =$ $\lim\limits_{x \to 3} (2(3)+1) = 7$ The value of $f(3) = 6$, but we discovered that $\lim\limits_{x \to 3} (2(3)+1) = 7$ so the function is discontinuous. To draw the graph we draw the function of $f(x) = (2x+1)$ and $(3,6)$.