Solutions

Take $f(x)=\sqrt{x}$ and consider its limit at $a=0$. We have

$$|x| < \delta = \epsilon^2 \implies |\sqrt{x}| < \epsilon.$$

However,

$$|x| < \frac{\delta}{2} = \frac{\epsilon^2}{2} \implies |\sqrt{x}| < \frac{\epsilon}{\sqrt{2}},$$

and $\frac{\epsilon}{2}<\frac{\epsilon}{\sqrt{2}}$, so we have looser guarantees on the range of $f(x)$.