## Calculus: Early Transcendentals 8th Edition

Not true. The counterexample is the function $$f(x)=1+e^{-\frac{1}{x^2}}.$$ This funtion is everywhere greater than $1$ but when $x\to 0$ we have $$\lim_{x\to0}(1+e^{-\frac{1}{x^2}})=\left[1+e^{-\frac{1}{0^+}}\right]=\left[1+e^{-\infty}\right]=1$$ which is not greater than one.