Question

if $sqrt{7 - 6x^{2}}leq f(x)leqsqrt{7 - x^{2}}$ for $- 1leq xleq1$, find $lim_{x
ightarrow0}f(x)$. $lim_{x
ightarrow0}f(x)=sqrt{7}$ (type an exact answer, using radicals as needed.)

Explanation:

Step1: Find left - hand limit

Find $\lim_{x
ightarrow0}\sqrt{7 - 6x^{2}}$. Substitute $x = 0$ into $\sqrt{7 - 6x^{2}}$, we get $\sqrt{7-6\times0^{2}}=\sqrt{7}$.

Step2: Find right - hand limit

Find $\lim_{x
ightarrow0}\sqrt{7 - x^{2}}$. Substitute $x = 0$ into $\sqrt{7 - x^{2}}$, we get $\sqrt{7 - 0^{2}}=\sqrt{7}$.

Step3: Apply Squeeze Theorem

Since $\sqrt{7 - 6x^{2}}\leq f(x)\leq\sqrt{7 - x^{2}}$ for $- 1\leq x\leq1$ and $\lim_{x
ightarrow0}\sqrt{7 - 6x^{2}}=\lim_{x
ightarrow0}\sqrt{7 - x^{2}}=\sqrt{7}$, by the Squeeze Theorem, $\lim_{x
ightarrow0}f(x)=\sqrt{7}$.

Answer:

$\sqrt{7}$