## Thomas' Calculus 13th Edition

The graph of $U(x)$ does not consist of a vertical tangent at the origin.
1. The graph of $U(x)$ has a vertical tangent at $x$ when $U(x)$ is continuous at $x$ 2. The graph of $U(x)$ has a vertical tangent at $x$ when $U'(x)=\lim\limits_{h\to 0}\dfrac{U(x+h)-U(x)}{h}=\pm\infty$ Also, $\lim\limits_{x\to0^+}U(x)=\lim\limits_{x\to0^+}(1)=1$ and $\lim\limits_{x\to0^{-}}U(x)=\lim\limits_{x\to0^{-}}(0)=0$ Since $\lim\limits_{x\to0^+}U(x)\ne\lim\limits_{x\to0^{-}}U(x)$, thus, $\lim\limits_{x\to 0} U(x)$ does not exist. This implies that $U(x)$ is not continuous at the origin and the graph of $U(x)$ does not consist of a vertical tangent at the origin.