Thomas' Calculus 13th Edition

$\sqrt5$
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 $\lim\limits_{x\to0}f(x)=\sqrt5$