Answer
True.
Work Step by Step
See p.76, The Precise Definition of a Limit:
...
$\displaystyle \lim_{x\rightarrow a}f(x)=L$
if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that
if $ 0 < |x-a| < \delta$ then $|f(x)-L| < \epsilon$.
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The statement matches the definition, with $\epsilon =1$,
$a=0, L=6.$
