#### Answer

$\sqrt5$

#### Work Step by Step

Here, we have
$\lim\limits_{x\to0}\sqrt{5-2x^2}=\sqrt{5-2\times0^2}=\sqrt{5-0}=\sqrt5$
and $\lim\limits_{x\to0}\sqrt{5-x^2}=\sqrt{5-0^2}=\sqrt5$
Then, $\lim\limits_{x\to0}\sqrt{5-2x^2}=\lim\limits_{x\to0}\sqrt{5-x^2}=\sqrt5$
As we can see that $\sqrt{5-2x^2}\le f(x)\le \sqrt{5-x^2}$ for $-1\le x\le 1$
Thus, as per Sandwich Theorem, we can see that
$\lim\limits_{x\to0}f(x)=\sqrt5$