Answer
$$\lim_{x\to0}f(x)=\sqrt5$$
Work Step by Step
- Calculate $\lim_{x\to0}\sqrt{5-2x^2}$ and $\lim_{x\to0}\sqrt{5-x^2}$
$\lim_{x\to0}\sqrt{5-2x^2}=\sqrt{5-2\times0^2}=\sqrt{5-0}=\sqrt5$
$\lim_{x\to0}\sqrt{5-x^2}=\sqrt{5-0^2}=\sqrt5$
So, $\lim_{x\to0}\sqrt{5-2x^2}=\lim_{x\to0}\sqrt{5-x^2}=\sqrt5$
- Yet $\sqrt{5-2x^2}\le f(x)\le \sqrt{5-x^2}$ for $-1\le x\le 1$
Therefore, applying the Sandwich Theorem, we conclude that $$\lim_{x\to0}f(x)=\sqrt5$$
