#### Answer

False.

#### Work Step by Step

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.