## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to1}e^{x^3-x}=1$$
$$A=\lim\limits_{x\to1}e^{x^3-x}$$ Since $e^{x^3-x}$ is an exponential function, it is continuous at any number in its domain, which is $R$. That means $A=\lim\limits_{x\to1}e^{x^3-x}=e^{1^3-1}=e^0=1$