## University Calculus: Early Transcendentals (3rd Edition)

(a) The graph is shown below. (b) $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=0$ (c) $\lim_{x\to1}f(x)=0$
(a) The graph is shown below. (b) Looking at the graph, as $x$ approaches $1$ from both the left and the right, $f(x)$ would arbitrarily get close to $0$. So, $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=0$ - Check algebraically: $\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(1-x^2)=1-1^2=0$ $\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(1-x^2)=1-1^2=0$ (c) Since $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=0$, as $x$ approaches $1$ from either side, $f(x)$ would still approach $0$. Therefore, $\lim_{x\to1}f(x)=0$