QuestionIf $\sqrt{5-2 x^{2}} \leq f(x) \leq \sqrt{5-x^{2}}$ for $-1 \leq x \leq 1$, find $\lim _{x \rightarrow 0} f(x)$. $$\lim _{x \rightarrow 0} f(x)=$$ $\square$ (Type an exact answer, using radicals as needed.)

Studdy Solution
Apply the squeeze theorem. Since $\sqrt{5 - 2x^2} \leq f(x) \leq \sqrt{5 - x^2}$ and both limits are $\sqrt{5}$ as $x \rightarrow 0$:
$$\lim_{x \to 0} f(x) = \sqrt{5}$$
Solution: $$\lim_{x \to 0} f(x) = \sqrt{5}$$