## Calculus 8th Edition

To build a counterexample, $x^{2}$ is nonnegative, $x^{2}+1 \geq 1$, $f(x)=x^{2}+1$will not satisfy the condition $f(x) > 1$, for all x because $f(0)=1,$ so we define a function piecewise, split at x=0, $f(x)=\left\{\begin{array}{ll} x^{2}+1 & \mathrm{i}\mathrm{f} x\neq 0\\ 1.1 & \mathrm{i}\mathrm{f} x=0 \end{array}\right.$ So, now, $f(x) > 1$ for all $x$, but $\displaystyle \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}(x^{2}+1)=1$, so the the statement is false.